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Precalculus Course — Transcript

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0:18This video introduces functions and their domains and ranges. A function is a correspondence
0:25between input numbers, usually the x values and output numbers, usually the y values,
0:32that sends each input number to exactly one output number. Sometimes a function is thought
0:38of as a rule or machine, in which you can feed in x values as input and get out y values
0:48as output. So, non mathematical example of a function might be the biological mother
0:54function, which takes as input and a person, and it gives us output their biological mother.
1:02This function satisfies the condition that each input number object person, in this case,
1:09get sent to exactly one output person. Because if you take any person, they just have one
1:16biological mother. So that rule does give you a function. But if I change things around,
1:23and just use the the mother function, which sends to each person, their mother, that's
1:30no longer gonna be a function, because there are some people who have more than one mother,
1:34right, they could have a biological mother and adopted mother, or a mother and stepmother
1:39or any number of situations. So since there's, there's at least some people who you would
1:44put it as input, and then you'd get like more than one possible output that violates this,
1:50this rule of functions, that would not be a function. Now, most of the time, we'll work
1:54with functions that are described with equations, not in terms of mothers. So for example, we
1:59can have the function y equals x squared plus one. This can also be written as f of x equals
2:06x squared plus one. Here, f of x is function notation that stands for the output value
2:12of y. Notice that this notation is not representing multiplication, we're not multiplying f by
2:20x, instead, we're going to be putting in a value for access input and getting out a value
2:26of f of x or y. For example, if we want to evaluate f of two, we're plugging in two as
2:34input for x either in this equation, or in that equation. Since F of two means two squared
2:42plus one, f of two is going to equal five. Similarly, f of five means I plug in five
2:51for x. So that's going to be five squared plus one, or 26. Sometimes it's useful to
2:58evaluate a function on a more complicated expression involving other variables. In this
3:03case, remember, the functions value on any expression, it's just what you what you get
3:08when you plug in that whole expression for x. So f of a plus three is going to be the
3:17quantity a plus three squared plus one, we could rewrite that as a squared plus six a
3:25plus nine plus one, or a squared plus six a plus 10. When evaluating a function on a
3:33complex expression, it's important to keep the parentheses when you plug in for x. That
3:39way, you evaluate the function on the whole expression. For example, it would be wrong
3:47to write f of a plus three equals a plus three squared plus one without the parentheses,
3:55because that would imply we were just squaring the three and not the whole expression, a
4:00plus three. Sometimes the function is described with a graph instead of an equation. In this
4:09example, this graph is supposed to represent the function g of x. Not all graphs actually
4:16represent functions. For example, the graph of a circle doesn't represent a function.
4:22That's because the graph of a circle violates the vertical line test, you can draw a vertical
4:27line, and I'll intersect the graph in more than one point. But our graph it left satisfies
4:33the vertical line test, any vertical line intersects the graph, and at most one point,
4:38that means is a function because every x value will have at most one y value that corresponds
4:44to it. Let's evaluate gf two, note that two is an x value. And we'll use the graph to
4:52find the corresponding y value. So I look for two on the x axis and find the point on
4:59the graph. If that has that x value, that's right here. Now I can look at the y value
5:06of that point looks like three, and therefore g of two is equal to three. If I try to do
5:13the same thing to evaluate g of five, I run into trouble. Five is an x value, I look for
5:19it on the x axis. But there's no point on the graph that has that x value. Therefore,
5:27g of five is undefined, or we can say it does not exist. The question of what x values and
5:34y values make sense for a function leads us to a discussion of domain and range. The domain
5:40of a function is all possible x values that make sense for that function. The range is
5:46the y values that makes sense for the function. In this example, we saw that the x value of
5:51five didn't have a corresponding y value for this function. So the x value of five is not
5:58in the domain of our function J. To find the x values in the domain, we have to look at
6:04the x values that correspond to points on the graph. One way to do that is to take the
6:09shadow or projection of the graph onto the x axis and see what x values are hit. It looks
6:18like we're hitting all x values, starting at negative eight, and continuing up to four.
6:25So our domain is the x's between negative eight, and for including those endpoints,
6:33or we can write this in interval notation as negative eight comma four with square brackets.
6:39To find the range of the function we look for the y value is corresponding to points
6:44on this graph, we can do that by taking the shadow or projection of the graph onto the
6:51y axis, we seem to be hitting our Y values from negative five, up through three. So our
7:04range is wise between negative five and three are an interval notation, negative five, three
7:13with square brackets. If we made a function that's described as an equation instead of
7:18a graph, one way to find the domain and range are to graph the function. But it's often
7:24possible to find the domain at least more quickly, by using algebraic considerations.
7:30We think about what x values it makes sense to plug into this expression, and what x values
7:34need to be excluded, because they make the algebraic expression impossible to evaluate.
7:39Specifically, to find the domain of a function, we need to exclude x values that make the
7:47denominator zero. Since we can't divide by zero, we also need to exclude x values that
7:55make an expression inside a square root sign negative. Since we can't take the square root
8:01of a negative number. In fact, we need to exclude values that make an expression inside
8:05any even root negative because we can't take an even root of a negative number, even though
8:11we can take an odd root like a cube root of a negative number. Later, when we look at
8:17logarithmic functions, we'll have some additional exclusions that we have to make. But for now,
8:22these two principles should handle all functions We'll see. So let's apply them to a couple
8:27examples. For the function in part A, we don't have any square root signs, but we do have
8:33a denominator. So we need to exclude x values that make the denominator zero. In other words,
8:40we need x squared minus 4x plus three to not be equal to zero. If we solve x squared minus
8:494x plus three equal to zero, we can do that by factoring. And that gives us x equals three
8:56or x equals one, so we need to exclude these values. All other x values should be fine.
9:03So if I draw the number line, I can put on one and three and just dig out a hole at both
9:09of those. And my domain includes everything else on the number line. In interval notation,
9:16this means my domain is everything from negative infinity to one, together with everything
9:22from one to three, together with everything from three to infinity. In the second example,
9:30we don't have any denominator to worry about, but we do have a square root sign. So we need
9:35to exclude any x values that make three minus 2x less than zero. In other words, we can
9:43include all x values for which three minus 2x is greater than or equal to zero. Solving
9:50that inequality gives us three is greater than or equal to 2x. In other words, x is
9:57less than or equal to three halves. If I can draw this on the number line, or write it
10:04in interval notation, notice that three halves is included. And that's because three minus
10:122x is allowed to be zero, I can take the square root of zero, that's just zero. And that's
10:16not a problem. Finally, let's look at a more complicated function that involves both the
10:22square root and his denominator. Now there are two things I need to worry about. I need
10:28the denominator to not be equal to zero, and I need the stuff inside the square root sign
10:36to be greater than or equal to zero. from our earlier work, we know the first condition
10:41means that x is not equal to three, and x is not equal to one. And the second condition
10:48means that x is less than or equal to three halves. Let's draw both of those conditions
10:53on the number line. x is not equal to three and x is not equal to one means we've got
11:00everything except those two dug out points. And the other condition x is less than or
11:07equal to three halves means we can have three halves, and everything to the left of it not
11:14to be in our domain. And to be legit for our function, we need both of these conditions
11:18to be true. So I'm going to connect these conditions with an and and that means we're
11:23looking for numbers on the number line that are colored both red and blue. So I'll draw
11:28that above and purple. So that's everything from three halves to one, I have to dig out
11:33one because one was a problem for the denominator. And then I can continue for all the things
11:38that are colored both both colors red and blue. So my final domain is going to be let's
11:45see negative infinity, up to but not including one together with one, but not including it
11:53to three halves, and I include three half since that was colored both red and blue.
11:58Also, in this video, we talked about functions, how to evaluate them, and how to find domains
12:03and ranges. This video is a brief introduction to increasing and decreasing functions based
12:10on their graphs. In this first example, I've graphed two lines, the first line is an increasing
12:19function. Because as the x values go from left to right, the y values are going up.
12:28this can be written more formally, by saying if x two is bigger than x one, these are supposed
12:34to be x values. So that means that x two is some x value, that's to the right so that
12:40it's bigger than x one. So whenever we have an x value x to this bigger than x one, increasing
12:47means that f of x two has to be bigger than f of x one, so that the y values are getting
12:57bigger when the x values are bigger. The second line is an example of a decreasing function.
13:07Because here as the x values increased from left to right, the y values are going down.
13:14So again, we can write this more formally by saying that whenever we have an x value
13:20x two that's bigger than x one, for example, this might be x two, and this might be x one,
13:26then the y value f of x two is less than f of x one this time, because here, when we
13:36have a bigger x value, we want to have a smaller y value to be decreasing. This second example
13:44gives the graph of a function that's increasing in some places, but decreasing and others.
13:50This part of the graph, here the functions decreasing, because as the x values increased
13:56from left to right, the y values go down. Over here, we also have a part of the graph
14:03where the function is decreasing. Now this part of the graph, the function is increasing,
14:09because as x goes from left to right, the y values are going up. I've marked that one
14:15in green, for increasing. And finally, on this part of the graph, the functions are
14:20neither increasing nor decreasing, it's completely flat or constant. Now we're asked to describe
14:25the intervals on which the graph is increasing or decreasing. So we describe those intervals
14:30in terms of the x values. It wouldn't make sense to describe them in terms of the y values
14:36because the y values can be the same for different parts of the graph, but the x values are always
14:40unique. Describe where this function is decreasing. That's for x values between negative four
14:51and negative two, and between four and seven. I can write this using inequalities as negative
14:58for less than x less than negative two, for less than x less than seven. It's not important
15:06whether I use less than or less than or equal to signs here, I'm going to use less than
15:11or equal to signs at the endpoints of the functions domain where the function stops
15:18existing altogether. And I'll use strict less than signs just where the function starts
15:24or stops decreasing in the middle of its domain. But again, that's not important. Describe
15:30where the function is increasing. That's for x values in between negative two, and one.
15:40So I can write that as an inequality as negative two less than x less than one, I can also
15:48describe these intervals in interval notation. So that would be close bracket, negative four,
15:54negative to open bracket, a cup sign for union, and then four comma seven, for the decreasing
16:01part, and the increasing part is negative two, comma one, I'm going to modify my graph
16:08just a little bit. Now by putting arrows on the end, what I have arrows instead of dots
16:13are just hard stops, that signifies that the function continues in the same direction forever.
16:20So if I write it like that, then the place where the function is increasing is still
16:25the same, it's still going to be from x values going from negative two to one. But now the
16:33decreasing part of the function extends further. So this section, these x values extend all
16:40the way out to infinity in it, we assume that it keeps going keeps decreasing here. So I
16:45would write that as for infinity, and similar the the left part, I would write as negative
16:55infinity, negative two, because the function is going on forever in this direction to set
17:00x values go all the way to negative infinity on that chunk. So when talking about increasing
17:06and decreasing parts of functions based on graphs, the thing to remember is that x is
17:13going to be heading from left to right. As long as you do that, increasing just means
17:20that y goes up. Decreasing means that y goes down. And if you're after describing the intervals,
17:28you have to do that in terms of x values. This video is about identifying maximums and
17:38minimums for functions from their graphs. We'll start with some definitions. A function
17:46f of x has an absolute maximum at the x value of x equals C, if the y value at x equals
17:53C is as big as it ever gets. We can write this more precisely by saying that f of c,
18:03that's the y value at x equals C is bigger than or equal to f of x, that's the y value
18:09at some other x value for all x values in the domain of f, that biggest y value f a
18:22c is called the absolute maximum value for F. And the point with an x value and a y value
18:32given is called an absolute maximum point. If I have a graph of a function, then the
18:40absolute maximum value, f of c is the highest value that function ever achieves. And the
18:48absolute maximum point is the point where it achieves that value. Now it's possible
18:55for a function to have more than one absolute maximum point, if there happens to be a tie
19:00where that highest value is achieved. But a function can only have at most one absolute
19:07maximum value. function f of x has an absolute minimum at x equals C. If the y value at x
19:20equals C is as small as it ever gets. We can write this more precisely by saying that f
19:28of c is less than or equal to f of x for all x values in the domain of f. the y value f
19:40of c is called the absolute minimum value of f. And the point c f of c is called an
19:52absolute minimum point in the graph of f of x. f of c is now The lowest y value that the
20:03graph ever achieves. And the point c f of c is the point where it achieves this very
20:08low y value. For example, this function has an absolute minimum value of about negative
20:17eight, and it has an absolute minimum point with coordinates, three, negative eight. If
20:26this function stops here, and has domain from x values just from zero to four, then the
20:33function will have an absolute maximum value of 10, achieved at the point with coordinates
20:42for 10. If however, the function keeps going on forever, then it will have no absolute
20:50maximum value at all. absolute maximum minimum values can also be called global maximum and
20:59minimum values. A function f of x has a local maximum at x equals C. If the y value at the
21:10x value of c is bigger than any y values nearby, it doesn't have to be the biggest y value
21:17anywhere in the domain. For example, for this function, there's bigger y values up here,
21:22it just has to be the biggest y value nearby, or more precisely, in an open interval of
21:29X values around C. We can write this by saying that f of c is bigger than or equal to f of
21:37x for all x values, and an open interval around C, the y value f OC is called a local maximum
21:52value for F. And the point with coordinates c f OC is called a local maximum point, a
22:01function f of x has a local minimum at x equals C. If f of c is less than or equal to f of
22:10x, for all x values in an open interval around the y value f of c is called a local minimum
22:21value. And the point with coordinate c f a c is called a local minimum point. A function
22:32might have many local minimum values. In this example, assuming that the domain is from
22:44zero to four, we have a local minimum point at x equals three, because this point here
22:53is the lowest point anywhere nearby. It also happens to be an absolute minimum point. Turning
23:01our attention to local maximum points. This function has a local maximum point at one
23:07to say, because this point here is the highest point anywhere nearby, even though it's not
23:14the highest point anywhere in the domain since their highest points higher points over here.
23:19So we have a local max point at one two, which is not an absolute max. different sources
23:30use different conventions as far as whether the point 410 up here counts as a local maximum
23:36point or not. Some sources do consider it a local maximum, since it's the highest point
23:41anywhere nearby. But other sources say that it doesn't count as a local maximum, for the
23:46technical reason that you can't get an open interval values around four because the function
23:53is not defined in an open interval around four because it's domain stops at four. So
23:58I'll just write this as a caution that some sources do not consider for 10 to be a local
24:13max point, because f is not defined. In an open interval around x equals four, local
24:29maximum and minimum values can also be called relative maximum and minimum values. Please
24:39pause the video for a moment to mark all local maximum minimum points on the graph of this
24:45function, as well as all absolute max and min points. The function has a local max point
24:52here at the point with coordinates approximately say 2.7 3.3 It's a local max point because
25:02it's the highest point anywhere nearby. There's a local min point right here at the point
25:09with coordinates to to, since this is the lowest point anywhere nearby. If this point
25:15actually existed, it would also be a local main points as well as an absolute min point.
25:21But this drilled out circle means that point is not actually part of the graph. So there's
25:26no local min point here, and no absolute min points. The point here with coordinates 04,
25:34is considered by some sources to be a local max point, but by other sources is not considered
25:39a local max point. Since the function is not defined in an open interval around zero, turning
25:45our attention to absolute maximum and points while there's no absolute max points as the
25:50function keeps going up and up and up forever. There's also no absolute min points, because
25:59the function just keeps going down and down a little by little and never actually achieves
26:04a lowest value. Now if we're talking about local maximum values, that's just the y values
26:11of the local maxima endpoints. So there's a local max value at this y value of 3.3.
26:17And there's a local min value of two this points y value. There are no absolute max
26:24or min values for this function. In this video, we defined absolute and local, max and min
26:36points and values. This video is about symmetry of graphs and functions. We say that a graph
26:44is symmetric with respect to the x axis, if the graph has mirror symmetry, with the x
26:54axis as the mirror line. This graph here is an example. Here is the mirror line. Notice
27:04that the point five three is on this graph. And so it's mirror image, five, negative three
27:13is also on the graph. Similarly, the point two, two is on the graph and its mirror image,
27:20two, negative two is on the graph. For any point x, y on a graph, its mirror image when
27:28you flip over the x axis will have coordinates the same x coordinate and the opposite of
27:36the y coordinates. Therefore, we can say that a graph is symmetric with respect to the x
27:41axis. If whenever a point x, y is on the graph, the point x negative y is also on the graph.
27:50We say that a graph is symmetric with respect to the y axis, if it has mirror symmetry,
27:59with the y axis as the mirror line. This graph is symmetric with respect to the y axis, here
28:07is the mirror line. Notice that the point with coordinates two three is on this graph.
28:16Its mirror image, which is also on the graph has coordinates negative two, three, so the
28:22same y coordinate, but the opposite x coordinate. And in fact, if I take any point with coordinates
28:30x, y, and I want its mirror image with the y axis as a mirror line, I'm going to get
28:37the point with the same height the same y coordinate, but the opposite x coordinate.
28:46So I can say that a graph is symmetric with respect to the y axis. If whenever a point
28:50x, y is on the graph, the point negative x, y is also on the graph. Finally, a graph is
28:58symmetric with respect to the origin if it has 180 degree rotational symmetry around
29:10the origin. This means that if I spin it by 180 degrees, the graph should line up exactly
29:20with itself. Rotating a graph by 180 degrees is the same thing as turning it upside down.
29:26So one way to see if a graph has symmetry respect to the origin, is to turn it upside
29:31down and see if it looks exactly the same. This one does. playing around with points,
29:38I noticed that the point one, negative two is on this graph. If I rotate that point around
29:46the origin by 180 degrees, I get to the point with coordinates negative one, two. If I start
29:56with a point on the graph with coordinates, two, three, And I rotate by 180 degrees, I
30:02get the point with coordinates negative two, negative three. And in general, if I start
30:10with a point x, y, and I rotate that 180 degrees around the origin, I get the point with coordinates
30:19negative x, negative y. so we can say that a graph is symmetric with respect to the origin.
30:25If whenever a point x, y is on the graph, the point negative x negative y is also on
30:31the graph. Please pause the video for a moment and decide which of these graphs are symmetric
30:36with respect to the x axis, the y axis and the origin. Some graphs may have more than
30:42one type of symmetry. Graph a is symmetric with respect to the origin. Because if you
30:49rotate it by 180 degrees, that is you turn it upside down, it looks exactly the same.
30:54It does not have any mirror symmetry, so it's not symmetric with respect to the x axis or
30:58the y axis. Graph B does have mirror symmetry, I'll draw on the two mirror lines. So it's
31:06both symmetric with respect to the x axis and symmetric with respect to the y axis.
31:13It also looks exactly the same upside down. So it's also symmetric with respect to the
31:17origin. Graph C is symmetric with respect to the y axis, but has no other symmetry.
31:25And graph D is symmetric with respect to the x axis with no other symmetry. We use the
31:33words even and odd to describe functions whose graphs have certain kinds of symmetry. A function
31:41is even if its graph is symmetric, with respect to the y axis. Remember that being symmetric
31:50with respect to the y axis means that whenever a point x, y is on the graph, its mirror image,
31:58negative x, y is also on the graph. That is, the y values corresponding to x and negative
32:07x are the same. Using function notation, I can write these points as the points with
32:15coordinates x, f of x and negative x, f of negative X, since the y value for a function
32:24is given by f of its x value. So we can say that a function f of x is even if F of negative
32:33X equals f of x, for all x values in its domain, the function has the same height at x and
32:41negative x. For example, we can see the function f of x equals x squared plus three is even.
32:49And one way to see this is by looking at its graph, which looks pretty much like the graph
32:52I've drawn. And that is saying that this graph is symmetric across the y axis. But we can
33:00also check that this function is even algebraically. without knowing anything about this graph,
33:04all we have to do is check that F of negative X is equal to f of x. So if we plug in at
33:12negative x for x into this formula, that's negative x squared plus three, but negative
33:19x squared is the same thing as x squared. And so this is exactly the same thing as f
33:25of x. And the property of being even is satisfied. We say that a function f of x is odd, if its
33:35graph is symmetric with respect to the origin. Recall that a graph is symmetric with respect
33:42to the origin means that whenever a point x, y is on the graph, it's 180 80 degree rotated
33:51point, negative x, negative y is also on the graph. That is the graph y value at x. And
34:00the graphs y value at negative x are the opposites of each other, same magnitude but opposite
34:07sign. If we use function notation, for the coordinates of these points, this one is x,
34:13f of x, and this one is negative x, f of negative x, right because the y value for a function
34:23at the x value of negative x is just F of negative X. So for a function to be odd, we
34:29see that F of negative X, that y value has to be the opposite of f of x, the other guys
34:38y value. A function f of x is odd if F of negative X is the negative of f of x for all
34:48x in its domain. Let's look at this example. We could draw its graph and verify the functions
34:56odd by seeing that rotational symmetry around the order On this graph, but we can also figure
35:01it out algebraically by checking to see if F of negative X is equal to negative of f
35:08of x. So let's plug in negative x for x in the formula for F, that gives us five times
35:14negative x minus one over negative x, that simplifies to negative 5x plus one over x.
35:22And we can see that that is actually the negative 5x minus one over x by factoring out a negative
35:30sign, which is negative f of x. So the property for being odd holds. In this video, we talked
35:41about symmetry with respect to the x axis symmetry with respect to the y axis, and symmetry
35:49with respect to the origin. functions that are symmetric with respect to the y axis are
35:55even functions and functions that are symmetric with respect to the origin, or odd functions.
36:01There's no word for functions that are symmetric with respect to the x axis. And that's because
36:05if your graph is symmetric with respect to the x axis, it's not going to be a function.
36:11This video gives the graphs of some commonly used functions that I call the toolkit functions.
36:17The first function is the function y equals x, let's plot a few points on the graph of
36:23this function. If x is zero, y zero, if x is one, y is one, and so on y is always equal
36:32to x doesn't have to just be an integer, it can be any real number, and we'll connecting
36:38the dots we get a straight line through the origin. Let's look at the graph of y equals
36:44x squared. If x is zero, y is zero. So we'll go through the origin again. If x is one,
36:51y is one, and x is negative one, y is also one, the x value of two gives a y value of
37:01four and the x value of negative two gives a y value of four also connecting the dots,
37:07we get a parabola. That is this, this function is an even function. That means it has mirror
37:17symmetry across the y axis, the left side it looks like exactly like the mirror image
37:22of the right side. That happens because when you square a positive number, like two, you
37:28get the exact same y value as when you square it's a mirror image x value of negative two.
37:36The next function y equals x cubed. I'll call that a cubic. Let's plot a few points when
37:42x is zero, y is zero. When x is one, y is one, when x is negative one, y is negative
37:50one, two goes with the point eight way up here, and an x value of negative two is going
37:56to give us negative eight. If I connect the dots, I get something that looks kind of like
38:01this. This function is what's called an odd function, because it has 180 degree rotational
38:11symmetry occur around the origin. If I rotate this whole graph by 180 degrees, or in other
38:16words, turn the paper upside down, I'll get exactly the same shape. The reason this function
38:23has this odd symmetry is because when I cube a positive number, and get its y value, I
38:30get n cube the corresponding negative number to get its y value, the negative number cubed
38:36gives us exactly the negative of the the y value we get with cubing the positive number.
38:43Let's look at the next example. Y equals the square root of x. Notice that the domain of
38:50this function is just x values bigger than or equal to zero because we can't take the
38:56square root of a negative number. Let's plug in a few x values. X is zero gives y is 0x
39:03is one square root of one is one, square root of four is two, and connecting the dots, I
39:10get a function that looks like this. The absolute value function is next. Again, if we plug
39:17in a few points, x is zero goes with y equals 0x plus one gives us one, the absolute value
39:25of negative one is one to two is on the graph and the absolute value of negative two is
39:32to I'm ending up getting a V shaped graph. It also has even or a mirror symmetry. Y equals
39:43two to the x is what's known as an exponential function. That's because the variable x is
39:49in the exponent. If I plot a few points, two to the zero is one Two to the one is to two
40:02squared is four, two to the minus one is one half. I'll plot these on my graph. Let me
40:10fill in a few more points. So let's see, two cubed is eight, that's way up here. And negative
40:17two gives me 1/4 1/8. Connecting the Dots, I get something that's shaped like this. You
40:26might have heard the expression exponential growth, when talking about, for example, population
40:30growth, this is function is represents exponential growth, it grows very, very steeply. Every
40:37time we increase the x coordinate by one we double the y coordinate. We could also look
40:46at a function like y equals three dx, or sometimes y equals e to the x, where E is just a number
40:53about 2.7. These functions look very similar. It's just the bigger base makes us rise a
41:01little more steeply. Now let's look at the function y equals one over x. It's not defined
41:08when x is zero, but I can plug in x equals one half whenever one half is two. Whenever
41:17one is one, and one over two is one half, by connect the dots, I get this shape in the
41:25first quadrant, but I haven't looked at negative values of x, when x is negative, whenever
41:30negative one is negative one, whenever negative half is negative two, and I get a similar
41:37looking piece in the third quadrant. This is an example of a hyperbola. And it's also
41:47an odd function, because it has that 180 degree rotational symmetry, if I turn the page upside
41:54down, it'll look exactly the same. Finally, let's look at y equals one over x squared.
42:00Again, it's not defined when x is zero, but I can plug in points like one half or X, let's
42:08see one over one half squared is one over 1/4, which is four. And one over one squared,
42:17one over two squared is a fourth, it looks pretty much like the previous function is
42:23just a little bit more extreme rises a little more steeply, falls a little more dramatically.
42:28But for negative values of x, something a little bit different goes on. For example,
42:33one over negative two squared is just one over four, which is a fourth. So I can plot
42:39that point there, and one over negative one squared, that's just one. So my curve for
42:47negative values of x is gonna lie in the second quadrant, not the third quadrant. This is
42:52an example of an even function because it has perfect mirror symmetry across the y axis.
43:01These are the toolkit functions, and I recommend you memorize the shape of each of them. That
43:08way, you can draw at least a rough sketch very quickly without having to plug in points.
43:14That's all for the graphs of the toolkit functions. If we change the equation of a function, then
43:22the graph of the function changes or transforms in predictable ways. This video gives some
43:29rules and examples for transformations of functions. To get the most out of this video,
43:33it's helpful if you're already familiar with the graph of some typical functions, I call
43:38them toolkit functions, like y equals square root of x, y equals x squared, y equals the
43:44absolute value of x and so on. If you're not familiar with those graphs, I encourage you
43:49to watch my video called toolkit functions first, before watching this one. I want to
43:55start by reviewing function notation. If g of x represents the function, the square root
44:02of x, then we can rewrite these expressions in terms of square roots. For example, g of
44:08x minus two is the same thing as the square root of x minus two. g of quantity x minus
44:17two means we plug in x minus two everywhere we see an x, so that would be the same thing
44:25as the square root of quantity x minus two. In this second example, I say that we're subtracting
44:33two on the inside of the function, because we're subtracting two before we apply the
44:38square root function. Whereas on the first example, I say that the minus two is on the
44:43outside of the function. We're doing the square root function first and then subtracting two.
44:48In this third example, g of 3x. We're multiplying by three on the inside of the function. To
44:54evaluate this in terms of square root, we plug in the entire 3x for x, and this square
45:00root function, that gives us the square root of 3x. In the next example, we're multiplying
45:07by three on the outside of the function, this is just three times the square root of x.
45:12Finally, g of minus x means the square root of minus x. This might look a little odd,
45:20because we're not used to taking the square root of a negative number. But remember that
45:24if x itself is negative, like negative to the negative x will be negative negative two
45:29or positive two. So we're really be taking the square root of a positive number in that
45:33case, let me record which of these are inside and which of these are outside of my function.
45:40In this next set of examples, we're using the same function g of x squared of x. But
45:45this time, we're starting with an expression with square roots in it, and trying to write
45:48it in terms of g of x. So the first example, the plus 17 is on the outside of my function,
45:56because I'm taking the square root of x first, and then adding 17. So I can write this as
46:02g of x plus 17. In the second example, I'm taking x and adding 12 first, then I take
46:11the square root of the whole thing. Since I'm adding the 12 to x first that's on the
46:15inside of my function. So I write that as g of the quantity x plus 12. Remember that
46:22this notation means I plug in the entire x plus 12, into the square root sign, which
46:28gives me exactly square root of x plus 12. And this next example of doing the square
46:33root first and then multiplying by negative 36, so i minus 36 multiplications, outside
46:39my function, I can rewrite this as minus 36 times g of x. Finally, in this last example,
46:47I take x multiplied by a fourth and then apply the square root of x. So that's the same thing
46:52as g of 1/4 x by 1/4. x is on the inside of my function. In other words, it's inside the
47:00parentheses when I use function notation. Let's graph the square root of x and two transformations
47:07of this function, y equals the square root of x goes to the point 0011. And for two,
47:13since the square root of four is two, it looks something like this. In order to graph y equals
47:21the square root of x minus two, notice that the minus two is on the outside of the function,
47:27that means we're going to take the square root of x first and then subtract two. So
47:32for example, if we start with the x value of zero, and compute the square root of zero,
47:36that's zero, then we subtract two to give us a y value of negative two, an x value of
47:42one, which under the square root function had a y value of one now has a y value that's
47:49decreased by two, one minus two is negative one. And finally, an x value of four, which
47:56under the square root function had a y value of two now has a y value of two minus two
48:02or zero, its y value is also decreased by two. If I plot these new points, zero goes
48:09with negative two, one goes with negative one, and four goes with zero, I have my transformed
48:16graph. Because I subtracted two on the outside of my function, my y values were decreased
48:23by two, which brought my graph down by two units. Next, let's look at y equals the square
48:30root of quantity x minus two. Now we're subtracting two on the inside of our function, which means
48:36we subtract two from x first and then take the square root. In order to get the same
48:40y value of zero as we had in our blue graph, we need our x minus two to be zero, so we
48:46need our original x to be two. In order to get the y value of one that we had in our
48:52blue graph, we need to be taking the square root of one, so we need x minus two to be
48:58one, which means that we need to start with an x value of three. And in order to reproduce
49:04our y value of two from our original graph, we need the square root of x minus two to
49:10be two, which means we need to start out by taking the square root of four, which means
49:16our x minus two is four, so our x should be up there at six. If I plot my x values, with
49:25my corresponding y values of square root of x minus two, I get the following graph. Notice
49:31that the graph has moved horizontally to the right by two units. To me, moving down by
49:39two units, makes sense because we're subtracting two, so we're decreasing y's by two units,
49:45but the minus two on the inside that kind of does the opposite of what I expect, I might
49:50expect it to to move the graph left I might expect the x values to be going down by two
49:55units, but instead, it moves the graph to the right because As the x units have to go
50:01up by two units, in order to get the right square root, when I then subtract two units,
50:07again, the observations we made for these transformations of functions hold in general,
50:12according to the following rules. First of all numbers on the outside of the function,
50:18like in our example, y equals the square root of x minus two, those numbers affect the y
50:24values, and result in vertical motions, like we saw, these motions are in the direction
50:29you expect. So subtracting two was just down by two. If we were adding two instead, that
50:35would move us up by two numbers on the inside of the function. That's like our example,
50:42y equals the square root of quantity x minus two, those affect the x values and results
50:47in a horizontal motion, these motions go in the opposite direction from what you expect.
50:53Remember, the minus two on the inside actually shifted our graph to the right, if it had
50:58had a plus two on the inside, that would actually shift our graph to the left. Adding results
51:05in a shift those are called translations, but multiplying like something like y equals
51:12three times a squared of x, that would result in a stretch or shrink. In other words, if
51:19I start with the square root of x, and then when I graph y equals three times the square
51:25root of x, that stretches my graph vertically by a factor of three. Like this, if I want
51:33to graph y equals 1/3, times the square root of x, that shrinks my graph vertically by
51:39a factor of 1/3. Finally, a negative sign results in reflection. For example, if I start
51:47with a graph of y equals the square root of x, and then want to graph y equals the square
51:52root of negative x, that's going to do a reflection in the horizontal direction because the negative
51:58is on the inside of the square root sign. A reflection in the horizontal direction means
52:04reflection across the y axis. If instead I want to graph y equals negative A squared
52:10of x, that negative sign on the outside means a vertical reflection, a reflection across
52:16the x axis. Pause the video for a moment and see if you could describe what happens in
52:23these four transformations. In the first example, we're subtracting four on the outside of the
52:30function. adding or subtracting means a translation or shift. And since we're on the outside of
52:36the function affects the y value, so that's moving us vertically. So this transformation
52:41should take the square root of graph and move it down by four units, that would look something
52:48like this. In the next example, we're adding 12 on the inside, that's still a translation.
52:57But now we're moving horizontally. And so since we go the opposite direction, we expect
53:02we are going to go to the left by 12 units, that's going to look something like this.
53:09And the next example, we're multiplying by three and introducing a negative sign both
53:13on the outside of our function outside our function means we're affecting the y values.
53:19So and multiplication means we're stretching by a factor of three, the negative sign means
53:24we reflect in the vertical direction, here's stretching by a factor of three vertically,
53:31before I apply the minus sign, and now the minus sign reflects in the vertical direction.
53:37Finally, in this last example, we're multiplying by one quarter on the inside of our function,
53:44we know that multiplication means stretch or shrink. And since we're on the inside,
53:49it's a horizontal motion, and it does the opposite of what we expect. So instead of
53:54shrinking by a factor of 1/4, horizontally, it's actually going to stretch by the reciprocal,
54:00a factor of four horizontally. that'll look something like this. Notice that stretching
54:07horizontally by a factor of four looks kind of like shrinking vertically by a factor of
54:12one half. And that's actually borne out by the algebra, because the square root of 1/4
54:18x is the same thing as the square root of 1/4 times the square root of x, which is the
54:23same thing as one half times the square root of x. And so now we can see algebraically
54:29that vertical shrink by a factor of one half is the same as a horizontal stretch by a factor
54:35of four, at least for this function, the square root function. This video gives some rules
54:42for transformations of functions, which I'll repeat below. numbers on the outside correspond
54:49to changes in the y values or vertical motions. numbers on the inside of the function, affect
54:58the x values and raise out in horizontal motions, adding and subtracting corresponds to translations
55:07or shifts. multiplying and dividing by numbers corresponds to stretches and shrinks. And
55:16putting in a negative sign. correspond corresponds to a reflection, horizontal reflection if
55:25the negative sign is on the inside, and a vertical reflection if the negative sign is
55:29on the outside. Knowing these basic rules about transformations empowers you to be able
55:35to sketch graphs of much more complicated functions, like y equals three times the square
55:41root of x plus two, by simply considering the transformations, one at a time. This video
55:48introduces piecewise functions. A piecewise function is a function whose defined in pieces
55:55by two or more different rules that apply for different x values. The rule for calculating
56:02this function is you calculate negative x squared if x is less than one, but you calculate
56:08negative 2x plus three if x is greater than or equal to one. So if we want to find f of
56:14negative two, well, negative two is less than one. So the first rule applies. And we compute
56:22f of negative two by plugging in negative two for x in the first rule, so that's negative
56:28four. Next, we want to find f of one. Well, one is right on the border in between the
56:36two rules. But because we have a greater than or equal to here, when x is equal to one,
56:42we apply this rule. And so we could plug one into the formula, negative 2x plus three sets
56:49negative two times one plus three, which gives us one as our output. Finally, if we want
56:54to compute f of three, since three is bigger than or equal to one, the second rule applies.
57:00And we plug three into that role. That gives us an answer of negative three. To graph F,
57:09it makes sense to also draw the graph in pieces. First, I'm going to draw the graph of minus
57:14x squared, just x squared would be a problem opening up, so minus x squared is a parabola
57:21opening down goes through the points negative one, negative one and one negative one. So
57:30it looks something like this. Now I've drawn the whole parabola, but the rule actually
57:37only applies when x is less than one. So I'll keep the part of the parabola when x is less
57:42than one. And I'll erase the part of the parabola when x is greater than or equal to one. I'll
57:49leave an open circle here, when x equals one since that point is not included in this definition,
57:55either. Next, I'm going to draw the second piece, the line y equals minus 2x plus three.
58:02So that's a line with slope negative two, and intercept y intercept three, so it goes
58:07through this point 03. And then it goes over by one and down by two. So it goes through
58:14the point one, one, over by one down by two. And
58:26I can continue and draw this straight line. Again, we only want part of this graph the
58:28part where x is greater than or equal to one. So I'll erase the part where x is less than
58:33one, this part here. This time, I'm going to leave a closed circle, where x is equal
58:39to one, since that point is included on that graph. When drawing this graph, it's good
58:44to think about what happens when x equals one the border point, both in the second role,
58:51where we actually include that point. And in the first rule, where we don't actually
58:55include that point, but we kind of have to draw it in order anyway to draw the open circle
58:59here. Now the last question asks us if this function is continuous. The informal definition
59:06of continuity that I'm going to use is that a function is continuous if you can draw the
59:09whole thing without picking up your pencil. And this case, we can't, because we have to
59:15pick up our pencil to get from the jump here, up to here. So this function is not continuous,
59:21it's got a discontinuity when x is one, which is exactly the border port between the two
59:26rules. Often piecewise functions will have discontinuities, where they transition from
59:32one role to the next. However, it's possible to have a piecewise defined function that
59:37has no discontinuity if the two pieces happen to line up perfectly. For example, if we've
59:43changed the functions definition slightly, I'll call it g of x this time to still be
59:48the negative x squared when x is less than one, but this time, negative 2x plus one,
59:54if x is bigger than equal to one, then when we graph it, The parabola piece will look
1:00:01the same, but the linear piece will be two units lower than before. And so it'll actually
1:00:08start right here at one, negative one and go down, and our function will be continuous.
1:00:16That's all for this introduction to piecewise functions. The inverse of a function undoes
1:00:22what the function does. So the inverse of tying your shoes would be to untie them. And
1:00:29the inverse of the function that adds two to a number would be the function that subtracts
1:00:36two from a number. This video introduces inverses and their properties. Suppose f of x is a
1:00:44function defined by this chart. In other words, f of two is three, f of three is five, f of
1:00:50four is six, and f of five is one, the inverse function for F written f superscript. Negative
1:00:591x undoes what f does. Since f takes two to three, F inverse takes three back to two.
1:01:10So we write this f superscript. Negative one of three is to. Similarly, since f takes three
1:01:19to five, F inverse takes five to three. And since f takes four to six, f inverse of six
1:01:28is four. And since f takes five to one, f inverse of one is five. I'll use these numbers
1:01:37to fill in the chart. Notice that the chart of values when y equals f of x and the chart
1:01:44of values, when y equals f inverse of x are closely related. They share the same numbers,
1:01:50but the x values for f of x correspond to the y values for f inverse of x, and the y
1:01:57values for f of x correspond to the x values for f inverse of x. That leads us to the first
1:02:04key fact inverse functions reverse the roles of y and x. I'm going to plot the points for
1:02:11y equals f of x in blue. Next, I'll plot the points for y equals f inverse of x in red.
1:02:20Pause the video for a moment and see what kind of symmetry you observe in this graph.
1:02:24How are the blue points related to the red points, you might have noticed that the blue
1:02:29points and the red points are mirror images over the mirror line, y equals x. So our second
1:02:38key fact is that the graph of y equals f inverse of x can be obtained from the graph of y equals
1:02:44f of x by reflecting over the line y equals x. This makes sense, because inverses, reverse
1:02:53the roles of y and x. In the same example, let's compute f inverse of f of two, this
1:03:01open circle means composition. In other words, we're computing f inverse of f of two, we
1:03:10compute this from the inside out. So that's f inverse of three. Since F of two is three,
1:03:19and f inverse of three, we see as to similarly, we can compute f of f inverse of three. And
1:03:31that means we take f of f inverse of three. Since f inverse of three is two, that's the
1:03:38same thing as computing F of two, which is three. Please pause the video for a moment
1:03:47and compute these other compositions. You should have found that in every case, if you
1:03:56take f inverse of f of a number, you get back to the very same number you started with.
1:04:02And similarly, if you take f of f inverse of any number, you get back to the same number
1:04:07you started with. So in general, f inverse of f of x is equal to x, and f of f inverse
1:04:15of x is also equal to x. This is the mathematical way of saying that F and envir f inverse undo
1:04:22each other. Let's look at a different example. Suppose that f of x is x cubed. Pause the
1:04:29video for a moment and guess what the inverse of f should be. Remember, F inverse undoes
1:04:35the work that F does. You might have guessed that f inverse of x is going to be the cube
1:04:43root function. We can check that this is true by looking at f of f inverse of x, that's
1:04:51F of the cube root of function, which means the cube root function cubed, which gets us
1:04:57back to x. Similarly, If we compute f inverse of f of x, that's the cube root of x cubed.
1:05:06And we get back to excellence again. So the cube root function really is the inverse of
1:05:12the cubing function. When we compose the two functions, we get back to the number that
1:05:17we started with. It'd be nice to have a more systematic way of finding inverses of functions
1:05:24besides guessing and checking. One method uses the fact that inverses, reverse the roles
1:05:31of y and x. So if we want to find the inverse of the function, f of x equals five minus
1:05:37x over 3x, we can write it as y equals five minus x over 3x. Reverse the roles of y and
1:05:46x to get x equals five minus y over three y, and then solve for y. To solve for y, let's
1:05:57multiply both sides by three y. Bring all terms with y's in to the left side, and alternate
1:06:06without y's and then to the right side, factor out the y and divide to isolate y. This gives
1:06:19us f inverse of x as five over 3x plus one. Notice that our original function f and our
1:06:28inverse function, f inverse are both rational functions, but they're not the reciprocals
1:06:33of each other. And then General, f inverse of x is not usually equal to one over f of
1:06:40x. This can be confusing, because when we write two to the minus one, that does mean
1:06:48one or two, but f to the minus one of x means the inverse function and not the reciprocal.
1:06:56It's natural to ask if all functions have inverse functions, that is for any function
1:07:02you might encounter. Is there always a function that it's it is its inverse? In fact, the
1:07:09answer is no. See, if you can come up with an example of a function that does not have
1:07:16an inverse function. The word function here is key. Remember that a function is a relationship
1:07:24between x values and y values, such that for each x value in the domain, there's only one
1:07:33corresponding y value. One example of a function that does not have an inverse function is
1:07:42the function f of x equals x squared. To see that, the inverse of this function is not
1:07:51a function. Note that for the x squared function, the number two and the number negative two,
1:08:00both go to number four. So if I had an inverse, it would have to send four to both two and
1:08:08negative two, the inverse would not be a function, it might be easier to understand the problem,
1:08:16when you look at a graph of y equals x squared. Recall that inverse functions reverse the
1:08:24roles of y and x and flip the graph over the line y equals x. But when I flip the green
1:08:31graph over the line y equals x, I get this red graph. This red graph is not the graph
1:08:37of a function, because it violates the vertical line test. The reason that violates the vertical
1:08:43line test is because the original green function violates the horizontal line test, and has
1:08:512x values with the same y value. In general, a function f has an inverse function if and
1:08:59only if the graph of f satisfies the horizontal line test, ie every horizontal line intersects
1:09:04the graph. And at most one point, pause the video for a moment and see which of these
1:09:10four graphs satisfy the horizontal line test. In other words, which of the four corresponding
1:09:16functions would have an inverse function you may have found that graphs A and B violate
1:09:25the horizontal line test. So their functions would not have inverse functions. But graphs
1:09:31C and D satisfy the horizontal line test. So these graphs represent functions that do
1:09:38have inverses. functions that satisfy the horizontal line test are sometimes called
1:09:44One to One functions. Equivalently a function is one to one if for any two different x values,
1:09:54x one and x two. the y value is f of x one and f of x two are done. different numbers.
1:10:01Sometimes as I said, f is one to one, if, whenever f of x one is equal to f of x two,
1:10:10then x one has to equal x two. As our last example, let's try to find P inverse of x,
1:10:17where p of x as the square root of x minus two drawn here, if we graph P inverse on the
1:10:24same axis as p of x, we get the following graph, simply by flipping over the line y
1:10:31equals x. If we try to solve the problem, algebraically, we can write y equal to a squared
1:10:38of x minus two, reverse the roles of y and x and solve for y by squaring both sides and
1:10:47adding two. Now if we were to graph y equals x squared plus two, that would look like a
1:10:55parabola, it would look like the red graph we've already drawn together with another
1:11:03arm on the left side. But we know that our actual inverse function consists only of this
1:11:10right arm, we can specify this algebraically by making the restriction that x has to be
1:11:20bigger than or equal to zero. This corresponds to the fact that on the original graph, for
1:11:28the square root of x, y was only greater than or equal to zero. Looking more closely at
1:11:35the domain and range of P and P inverse, we know that the domain of P is all values of
1:11:44x such that x minus two is greater than or equal to zero. Since we can't take the square
1:11:48root of a negative number. This corresponds to x values being greater than or equal to
1:11:54two, or an interval notation, the interval from two to infinity. The range of P, we can
1:12:02see from the graph is our y value is greater than or equal to zero, or the interval from
1:12:07zero to infinity. Similarly, based on the graph, we see the domain of P inverse is x
1:12:16values greater than or equal to zero, the interval from zero to infinity. And the range
1:12:21of P inverse is Y values greater than or equal to two, or the interval from two to infinity.
1:12:29If you look closely at these domains and ranges, you'll notice that the domain of P corresponds
1:12:35exactly to the range of P inverse, and the range of P corresponds to the domain of P
1:12:41inverse. This makes sense, because inverse functions reverse the roles of y and x. The
1:12:50domain of f inverse of x is the x values for F inverse, which corresponds to the y values,
1:12:56or the range of F. The range of f inverse is the y values for F inverse, which correspond
1:13:04to the x values or the domain of f. In this video, we discussed five key properties of
1:13:12inverse functions. inverse functions, reverse the roles of y and x. The graph of y equals
1:13:22f inverse of x is the graph of y equals f of x reflected over the line y equals x. When
1:13:33we compose F with F inverse, we get the identity function y equals x. And similarly, when we
1:13:42compose f inverse with F, that brings x to x. In other words, F and F inverse undo each
1:13:48other. The function f of x has an inverse function if and only if the graph of y equals
1:14:01f of x satisfies the horizontal line test. And finally, the domain of f is the range
1:14:15of f inverse and the range of f is the domain of f inverse. These properties of inverse
1:14:24functions will be important when we study exponential functions and their inverses logarithmic
1:14:31functions. There are different ways of describing the size of an angle. This video explains
1:14:39how to convert between them. angles are commonly measured using either degrees or radians.
1:14:49If we're talking about degrees, then a full circle, by convention is 360 degrees. A half
1:14:58circle then would be 180 degrees. And if I wanted to do say, a third of a circle, that
1:15:06would be a third of 360 degrees or 120 degrees. If we're using radians instead, then going
1:15:14all the way around the circle is called two pi radians. The two pi was chosen because
1:15:23that's the circumference of a circle with a radius of one, a half circle is then half
1:15:31of two pi radians, which is pi radians. And if I wanted to do say, a quarter of a circle,
1:15:39that would be a quarter of my two pi radians, a quarter of two pi is pi over two. To convert
1:15:50between degrees and radians is handed to us the fact that 180 degrees corresponds to pi
1:15:57radians. Both of these represent half of a circle. Let's convert negative 135 degrees
1:16:06to radians. Now, you might be wondering, what does a negative angle even mean? Well, by
1:16:12convention, if we go in the counterclockwise direction, that's considered a positive angle.
1:16:21And if we go in the clockwise direction, that's considered a negative angle. To convert negative
1:16:30135 degrees to radians, I could multiply by pi radians per 180 degrees. Notice that the
1:16:40degrees cancel here, and I end up with negative 135 pi over 180 radians. That simplifies to
1:16:53negative three Pi over four radians. Or if I convert pi to a decimal, that's negative
1:17:012.3562 radians, up to four decimal places. But usually, if the answer is a simple fraction
1:17:12times pi, it's better to leave it in that form. Rather than convert it to a decimal.
1:17:17I can also go the other direction and convert radians to degrees, five pi over four radians.
1:17:27Now again, I we need to use the fact that pi radians is 180 degrees. But if I write
1:17:31it like this, with radians on the top and degrees on the bottom, then the radians won't
1:17:37cancel. So that doesn't work. Instead, so instead, I need to find by the 180 degrees
1:17:44over hi radians, so they got cancel the radians that way left radio cancel, and I get five
1:17:52pi times 180 over four pi degrees. Now the PI's also cancel, and this simplifies to 225
1:18:02degrees. Converting seven radians to degrees is similar, I need to multiply by 180 degrees
1:18:10over pi radians. If I work out seven times 180 over pi on my calculator, I get 401 point
1:18:19O seven degrees up to two decimal places. Notice that my answer a little more than 400
1:18:26is more than 360 degrees, so this is more than a full circle. That makes sense since
1:18:32seven radians is more than two pi radians. Since pi is itself a little more than three.
1:18:40Sometimes angles are given in terms of degrees, minutes and seconds, as in 32 degrees, 17
1:18:50minutes, and 25 seconds. A minute is defined as 1/60 of a degree. A second is defined as
1:19:011/60 of a minute, but since a minute is itself 1/60 of a degree This is 1/60 of 1/60 of a
1:19:11degree, or one over 3600 of a degree. You can also think of this by saying that they're
1:19:1960 minutes in one degree and they're 3600 seconds in one degree. Let's convert this
1:19:34number of degrees minutes and seconds to a decimal number of degrees. We know that we
1:19:40have 32 degrees plus 17 minutes plus 25 seconds. I'm going to copy down the 32 degrees, but
1:19:52I need to convert the minutes to degrees. I know that there's 60 minutes in one degree
1:20:00If I want the unit's to cancel, I better put the degrees on the top, so one degree corresponds
1:20:05to 60 minutes. Similarly, if I want to convert seconds to degrees 25 seconds, I know that
1:20:16there are 3600 seconds in one degree, I want the degrees to end up on the top, and I want
1:20:22the seconds on the bottom to cancel with a 25 seconds. So I need to write it this way.
1:20:29Okay, now I can write this as 32 degrees plus 17 over 60 degrees plus 25 over 3600 degrees.
1:20:38And because the minutes cancelled, and the seconds cancelled, if I add this all up on
1:20:46my calculator, I get 32.2903 degrees up to four decimal places, I can also go the other
1:20:55direction, and convert a decimal number of degrees into degrees, minutes and seconds.
1:21:02I'm starting out with 247 degrees plus a decimal number of degrees leftover. If I want to convert
1:21:120.3486 degrees, two minutes, well, I know that I have 60 minutes in one degree, this
1:21:22time, I want the minutes on the top and the degrees on the bottom, so that my degrees
1:21:28will cancel. So I just multiply my decimal by 60. And I end up with 20.916 minutes. Let
1:21:35me copy this down. And now I'm going to take the 0.916 minutes and convert those two seconds.
1:21:49So I know that they're 60 seconds in one minute. So canceling my units again, I just multiply
1:21:56by 60, which gives me 54.96 minutes, I can copy this down as is. Or I can round off my
1:22:10seconds and get 247 degrees, 20 minutes, and 55 seconds. converting between different ways
1:22:20of measuring angles is all about unit conversion. To convert between radians and degrees, we
1:22:26use the fact that pi radians corresponds to 180 degrees. To convert between minutes and
1:22:35seconds and degrees, we use the fact that 60 minutes corresponds to one degree, and
1:22:4260 seconds corresponds to one minute. If you have a circle, a piece of the circumference
1:22:51is called an arc. And a wedge of pi for that circle is called a sector. This video explains
1:23:02how to calculate the length of the arc and the area of the sector in terms of the angle
1:23:09and the radius of the circle. the circumference of a circle is given by the formula, the circumference
1:23:18equals two pi times the radius. Let's use that to solve the following problem. A circular
1:23:25pool has a radius of eight meters, find the arc length spanned by a central angle of 2.5
1:23:34radians. So this angle from the center of the circle is supposed to be 2.5 radians.
1:23:40I've drawn the angle is a little bit less than half the circle because half the circle
1:23:45would be pi radians, which is 3.141 radians. So I want to find the length of this arc.
1:23:54I know that the total circumference is going to be two pi times eight meters. But I just
1:24:01want a fraction of the circumference. The arc length I want is going to be given by
1:24:07the fraction of the circle that the angle makes times the circumference of the circle.
1:24:17The fraction of the circle that I want is given by the ratio of 2.5 radians over the
1:24:24total number of radians in the circle, which is two pi radians. Multiply that by there's
1:24:30a conference, which we said was two pi times eight meters. Notice that the two PI's cancel,
1:24:39and I'm left with just 2.5 times eight meters, which is 10 meters. In general, the arc length
1:24:50is related to the angle it spans by considering the arc length to be the fraction of the circle
1:24:57you get by taking the angle over The total angle the circle of two pi times the circumference,
1:25:04which is two pi r. Since the two PI's cancel as they did in our example, that means that
1:25:10arc length is given by theta times r. here, R is the radius of the circle, and theta is
1:25:20the angle, it's important to note that theta needs to be measured in radians, not degrees
1:25:27for this to work. That's because when I took this ratio, I was using two pi radians for
1:25:34the total measure of the circle. So it's important that theta also be in radians. The area of
1:25:40a circle is given by the formula, pi r squared, where r is the radius, let's use that fact
1:25:48to find the area of the sector of a circle of radius 10 meters. that spans an angle of
1:25:56pi over six radians, we know that the total area of the circle is given by pi times 10
1:26:06meters squared, or 100 pi meter squared. But we just want a fraction of that area to give
1:26:14us the area of our sector. So we want to take the fraction of the circle that the sector
1:26:23makes times the area of the circle. Well, the fraction of the circle is given by the
1:26:31angle of pi over six over the total angle in the whole circle of two pi. And well times
1:26:41that by the 100 pi meter squared, that simplifies to 100 pi over 12 meters squared, or 25 thirds
1:26:55pi meter squared as a decimal, that's 26.18 meter squared up to two decimal places. In
1:27:04general, if you want to find the area of a sector, you can take the fraction of the circle
1:27:13that the sector spans out times the area of the circle, pi r squared. Notice that the
1:27:21fraction of the circle is given by the angle that the sector makes in radians, divided
1:27:28by the two pi radians in the circle. Since the PI's cancel, we can simplify this formula
1:27:38by saying the area is given by theta over two times r squared, where theta is the angle
1:27:47of the sector in radians. Again, it's important that this angle be in radians, since that's
1:27:55how we're doing our fraction of our circle comparison. As usual, r represents the radius
1:28:02of the circle. Notice that if you don't have your angle in radians, if it's in degrees
1:28:08instead, not a big deal, because we can always convert from degrees to radians first, before
1:28:13using our formula. In this video, we saw that for our sector of angle theta, given in radians,
1:28:24the arc length is given by the formula theta times R. And the area of the sector is given
1:28:33by theta over two times r squared, where r is the radius of the circle. This video defines
1:28:42linear and radial speed for a rotating circle, and explains how to convert between them.
1:28:50Consider a spinning wheel. The angular speed is the angle it goes through in a unit of
1:28:57time, I can write that as angle per time, the units will be something like radians per
1:29:08second if the angle is measured in radians, and the time is measured in seconds, or I
1:29:12guess we could do degrees per minute, etc. The linear speed is the speed of a point on
1:29:21the rim of the wheel. So that is the distance a point on the rim of the wheel travel travels
1:29:33in a unit of time. I can think of that as distance per time. And so it has units of
1:29:42something like meters per second for example, or it could be feet per minute, etc. Suppose
1:29:51we have a ferris wheel with radius 20 meters. That's making one revolution every two minutes,
1:30:01we want to find its angular speed and the linear speed of a point on its rim. Now the
1:30:06Ferris wheel is going one revolution every two minutes. So that's one half revolution
1:30:14per minute, I want to, I want to find the angular speed, which is the angle it goes
1:30:21through in a unit of time. I've already got a unit of time in the denominator minutes,
1:30:27but I've got to somehow convert revolutions to angle. So my angular speed, one half revolutions
1:30:35per minute, one revolution, but that on the bottom, so the units will cancel out. One
1:30:43revolution is going through an angle of two pi radians. That means my angular speed is
1:30:51going to be one half times two pi radians per minute, or pi radians per minute. The
1:31:03linear speed, the speed of a point on the rim, well, the wheel is going pi radians per
1:31:12minute, I got to somehow convert radians into distance. So I know that when I go to pi radians,
1:31:23so that's all the way around the circle, I'm going all the way around the circumference
1:31:27and distance, so that would be two pi times the radius, or two pi times 20 meters. The
1:31:34two PI's cancel as do the radians, and I'm left with 20 pi meters per minute. Let's review
1:31:46how we got from angular speed to linear speed in order to get a more general formula. I'll
1:31:52call the angular speed, omega looks like a W, I'll call the linear speed v. In the previous
1:32:01problem, we found the linear speed v, by starting with the angular speed omega, and multiplying
1:32:09it by the circumference divided by two pi. That's because the point on the rim travels
1:32:17the whole circumference as it goes through an angle of two pi all the way around the
1:32:23circle. So I'll write that down here, circumference over two pi. But since I know that my circumference
1:32:33is given by the formula two pi r, and the two PI's cancel, as they did in the previous
1:32:40problem, that shows that the linear speed is the angular speed times r, where r is the
1:32:47radius of the circle. In this video, we defined linear speed and radial speed, and found that
1:32:57linear speed was radial speed times the radius of the circle. This video introduces the trig
1:33:04functions, sine, cosine, tangent, secant, cosecant, and cotangent. For right triangles.
1:33:17For a right triangle, with sides of length A, B, and C angle theta as drawn, we define
1:33:30sine of theta as the length of the opposite side over the hypothesis, the side that's
1:33:39opposite to our angle theta has measure a and I have partners is this side here with
1:33:44measure C. So that would be a oversee for this triangle. Cosine of theta is defined
1:33:54as the length of the adjacent side, over the length of hi partners. This side here is the
1:34:00side adjacent to theta. Of course, the hypothesis is also adjacent to theta, but it's special
1:34:06as a partner, so we don't think of it as the adjacent side. So that would be B oversea.
1:34:13tangent of theta is the opposite side length over the adjacent side length, so that would
1:34:18be a over b. The pneumonic to remember this is so toa, that's sine is opposite over hypotenuse.
1:34:33Cosine is adjacent over hypotenuse. And tangent is opposite over adjacent. In fact, there's
1:34:44a relationship between tangent and sine and cosine. Namely, tangent of theta is equal
1:34:52to sine of theta over cosine of theta. If you want to see why that's that's because
1:34:59some of theta over cosine of theta is given by a sine, which is opposite over hypothesis,
1:35:07divided by cosine, which is adjacent over hypotenuse. If we compute these fractions
1:35:15by flipping and multiplying, the high partners, length cancels, and we just get opposite over
1:35:25adjacent, which is, by definition, tangent of theta. There are three more trig functions
1:35:31that are defined in terms of sine, cosine and tangent. First of all, their sequence
1:35:37of theta, by definition, that's one over cosine of theta. So it's going to be one over the
1:35:45adjacent over the partners, which is the high partners over the adjacent, which in this
1:35:51triangle is C over B. cosecant of theta is defined as one over sine theta. So that's
1:36:00one over the opposite over the hypotenuse, which is the hypothesis over the opposite.
1:36:07And for this triangle, that's going to be C over a. Finally, cotangent of theta is defined
1:36:14as one over tan theta. So that's going to be one over opposite over adjacent, flip and
1:36:22multiply, I get adjacent over opposite, which in this case is b over a. So notice that the
1:36:32values for cotangent cosecant and secant are the reciprocals of the values for tangent,
1:36:41sine and cosine, respectively. Let's use these definitions to find the exact values of all
1:36:49six trig functions for the angle theta in this triangle. I'll start with a sine of theta.
1:36:56That's the opposite of our hypothesis. Well, for this angle theta, the opposite side is
1:37:03down here and has measured to the high partners has measured five, sine theta is two over
1:37:10five. Cosine theta is adjacent over hypotenuse, but I don't know the value of the side length.
1:37:20But fortunately, I can find it using Bagaran theorem says I have a right triangle here,
1:37:24I know that a squared, I'll call the side length A plus B squared, where b is this other
1:37:33leg of the triangle is equal to c squared, where c is the hypothesis. So here I have
1:37:39a squared plus two squared equals five squared, which means that a squared plus four equals
1:37:4525. A squared is 21. So A is plus or minus the square root of 21. But since I'm talking
1:37:52about the length of a side of a triangle, I can just use the positive answer. Returning
1:37:58to my computation of cosine theta, I can write it as adjacent, which is the square root of
1:38:0521. Over high partners, which is five, tangent theta is the opposite over the adjacent, so
1:38:15that's going to be two over the square root of 21. To compute secant of theta, that's
1:38:23one over cosine theta, so that's going to be one over square root of 21 over five, which
1:38:30is five over the square root of 21. The reciprocal of my cosine value. cosecant theta is one
1:38:41over sine theta, that's going to be the reciprocal of my sine, so five halves, and cotangent
1:38:47theta is one over tan theta, so it's going to be the reciprocal of my 10 value square
1:38:53root of 21. Over to finally, we'll do an application. So if we have a kite that's flying at an angle
1:39:02of elevation as the angle from the horizontal of 75 degrees, with a kite length string of
1:39:10100 meters, we want to find out how high it is I'll call the height Why? Well, we want
1:39:19to relate the known quantities this angle, and this hypothesis, to the unknown quantity,
1:39:27the unknown quantity is the opposite side of our triangle. So if we use sine of theta
1:39:34equals opposite of our hypothesis, then we can relate these known amounts sine of 75
1:39:42degrees to our unknown amount y, which is the opposite and our note amount of 100 meters.
1:39:49solving for y This gives that y is 100 meters times sine of 75 degrees. We can use a calculator
1:39:56to compute sine of 75 degrees. Be sure Use degree mode and not radian mode when you type
1:40:04in the 75. When I do the computation, I get a final answer of 96.59 meters up to two decimal
1:40:16places. Notice that we're ignoring the height of the person in this problem. To remember
1:40:22the definitions of the trig functions, you can use the pneumonic, so tau, and the fact
1:40:29that secant is the reciprocal of cosine cosecant the reciprocal of sine and cotangent the reciprocal
1:40:36of tangent. In this video, I'll use geometry to compute the sine and cosine of a 30 degree
1:40:42angle, a 45 degree angle and a 60 degree angle. One way to compute the sine of a 45 degree
1:40:51angle is to use a right triangle with a 45 degree angle. This particular right triangle
1:40:58has I have hotness of length one. Since all the angles of the triangle have to add up
1:41:05to 180 degrees, and we already have 90 degrees and 45 degrees, the remaining angle must also
1:41:13be 45 degrees. So we have an isosceles triangle with two sides the same length, I'll call
1:41:19that side length a. If we want sine of 45 degrees, let's use this 45 degree angle here,
1:41:26then sine is opposite over hypotenuse. So if I can figure out how long this side length
1:41:31is, I'll be able to compute sine of 45 degrees. Now the Pythagorean theorem says this side
1:41:37length squared plus that side length squared equals I have hot news squared. So we have
1:41:42that a squared plus a squared equals one squared. Alright, that is two A squared equals one.
1:41:49So a squared is one half, and a is plus or minus the square root of one half. Since we're
1:41:57talking about the length of sides of triangles, I can just use the positive square root it's
1:42:04customary to rewrite this as the square root of one over the square root of two, which
1:42:09is one over the square root of two, and then rationalize the denominator by multiplying
1:42:13the top and the bottom by the square root of two. That gives me square root of two in
1:42:18the numerator, and the square root of two squared in the denominator, which is the square
1:42:22root of two over two. So the side lengths are the square root of two over two. Now I
1:42:29can figure out the sine of 45 degrees, by computing the opposite over the hypotenuse.
1:42:36That's, I'm looking at this angle. So opposite is square root of two over two hypothesis
1:42:42one. So the sine of 45 degrees is the square root of two over two. Cosine of 45 degrees
1:42:52is adjacent over hypotenuse. That's this side length over this hypothesis. So that's the
1:42:59square root of two over to over one again, what would happen if instead of using this
1:43:06triangle with hypotony is one, we use this triangle, also a 4545 90 triangle, but with
1:43:12hypotony is five, not quite drawn to scale. Please pause the video for a moment and repeat
1:43:18the computation with this triangle. This time, I'll call the side length B. Pythagorean Theorem
1:43:26tells me b squared plus b squared equals five squared. So two, b squared equals 25. And
1:43:33b squared equals 25 over two, B is going to be the plus or minus the square root of 25
1:43:39over two, again, I can just use the positive version. And so B is the square root of 25
1:43:45over the square root of two, which is five over the square root of two, rationalizing
1:43:51the denominator, I get five root two over two. Now sign of my 45 degree angle is opposite
1:44:01over her partners, which is five square root of two over two divided by five. That simplifies
1:44:10to the square root of two over two as before, and a similar computation shows that cosine
1:44:16of 45 degrees is also square root of two over two as before. This makes sense, because sine
1:44:24and cosine are based on ratios of sides. And since these two triangles are similar triangles,
1:44:30they'll have the same ratios of sides. To find the sine and cosine of 30 degrees, let's
1:44:37use this 30 6090 right triangle with hypotony is one. If we double the triangle, we get
1:44:48an angle of 30 here, so a total angle of 60 degrees here, and this angle is also 60 degrees.
1:44:55So we have a 60 6060 triangle. That's an equilateral triangle. All side length the same. Since
1:45:02this side length has length one, this side length is also one. This entire side length
1:45:10is one, which means this short side of our original triangle has length one half. Going
1:45:17back to my original triangle, let's use the Pythagorean Theorem to find the length of
1:45:22his longer side x. Pythagorean theorem says x squared plus one half squared equals one
1:45:32squared. So x squared plus a fourth equals 1x squared is three fourths. And so x is plus
1:45:39or minus the square root of three fourths is the positive version and get the square
1:45:45root of three over the square root of four, which is the square root of three over two.
1:45:51Now using our original triangle, again, let's compute the sine of this 30 degree angle here.
1:45:59We know that sine of 30 degrees is opposite of our hypothesis. The opposite of this angle
1:46:07is one half and the hypothesis is one. So we get a sine of one half over one, which
1:46:14is one half cosine of 30 degrees is adjacent over hypotenuse. So that's the square root
1:46:22of three over two divided by one. To find sine of 60 degrees and cosine of 60 degrees,
1:46:29we can actually use the same green triangle and just focus on this upper corner angle
1:46:35of 60 degrees instead. So sine of 60 degrees. Opposite overhype hotness, but this time the
1:46:45opposite to this angle is the square root of three over two. Cosine of 60 degrees adjacent
1:46:55over hypotenuse gives us one half. I'll summarize the results in this table below. Notice that
1:47:04a 30 degree angle corresponds to pi over six radians since 30 degrees times pi over 180
1:47:14is pi over six. Similarly, 45 degrees corresponds to pi over four radians and 60 degrees corresponds
1:47:22to pi over three radians. I recommend that you memorize the three numbers one half, root
1:47:30two over two, and root three over two. And the fact that one half and root three over
1:47:35two go together. And root two over two goes with itself. From that information, it's not
1:47:45hard to reconstruct the triangles, you know that a 4545 90 triangle is an isosceles triangle,
1:47:53so it must have the side links where the same number goes with itself. And a 30 6090 triangle
1:48:02has one side length smaller than the other. So the smaller side must be one half and the
1:48:08larger side must be root three over two since root three is bigger than one and so root
1:48:13three over two is bigger than one half. Doing a visual check, you can easily fill in the
1:48:21angles, the smaller angle must be the 30 degree angle, and the larger one must be the 60 degree
1:48:28one. In this video, we computed the sine and cosine of three special angles 30 degrees,
1:48:3545 degrees, and 60 degrees. This video defined sine and cosine in terms of points on the
1:48:45unit circle, a unit circle is a circle with radius one. Up to now we defined sine and
1:48:53cosine and tangent in terms of right triangles. For example, to find sine of 14 degrees in
1:48:59theory, you could draw a right triangle with an angle of 14 degrees and then calculate
1:49:08the sign as the length of the opposite side over the length of by partners. But if we
1:49:18use this method to try to compute sine of 120 degrees, things go horribly wrong. When
1:49:26we draw this 120 degree angle, and this right angle, there's no way to complete this picture
1:49:31to get a right triangle. So instead, we're going to use a unit circle that is a circle
1:49:38of radius one. The figure below illustrates how right triangles and a unit circle are
1:49:44related. If you draw right triangles with larger of hypotenuse one with larger and larger
1:49:51angles, then the top vertex sweeps out part of a unit circle. Let's look at this relationship
1:49:58in more detail. In this figure, I've drawn a right triangle inside a unit circle. The
1:50:07hypothesis of the triangle is the radius of the circle, which is one, one vertex of the
1:50:14right triangle is at the origin. And another vertex of the right triangle is at the edge
1:50:20of the circle, I'm going to call the coordinates of that vertex A, B. Now the base of this
1:50:29right triangle has length a, the x coordinate, and the height of the right triangle is B,
1:50:37the y coordinate. If I use the right triangle definition of sine and cosine of theta, this
1:50:45right here is the angle theta, then cosine of theta is adjacent over hypotenuse, so that's
1:50:53a over one, or a. Notice that a also represents the x coordinate of this point on the unit
1:51:03circle, at angle theta from the x axis, I'll write that down. For a sine of theta, if I
1:51:12use the right triangle definition, that's opposite overhead partners. So b over one,
1:51:19which is just B. But B also represents the y coordinate of this point in the on the unit
1:51:27circle at angle theta. For tangent theta, if we use the right side triangle definition,
1:51:34its opposite over adjacent. So that's B over A, I can think of that as the y coordinate
1:51:41of the point over the x coordinate of the point. Now for angles theta, that can't be
1:51:47part of a right triangle, because they're too big, they're bigger than 90 degrees, like
1:51:52now I'll call this angle here theta, I can still use this idea of x and y coordinates
1:52:00to calculate the sine and cosine of theta. So if I just mark this point, on the end of
1:52:08this line at angle theta, if I mark that to have coordinates x and y, then cosine theta,
1:52:15I'm still going to define as the x coordinate of this point, sine theta as the y coordinate,
1:52:23and tangent theta as the ratio of the y coordinate over the x coordinate. When we use this unit
1:52:32circle definition, we always draw theta starting from the positive x axis and going counterclockwise.
1:52:45Let's use this unit circle definition to calculate sine, cosine and tangent of this angle fee.
1:52:51In our figure, we have a unit circle. And these numbers are supposed to represent the
1:52:58x&y coordinates of this point on the unit circle at the endpoint of this line segment,
1:53:06that lies at angle fee for the positive x axis sign fee is equal to the y coordinate
1:53:17cosine fee is equal to the x coordinate. And tangent of fee is given by the ratio of the
1:53:25two, which works out to negative 0.3639 up to four decimal places. This video gives a
1:53:36method for calculating sine cosine and tangent in terms of the unit circle. Starting from
1:53:42the positive x axis, you draw the angle theta going counterclockwise. You look at the coordinates
1:53:52of the point on the unit circle where that angle ends. And cosine of that angle theta
1:54:02is the x coordinate, sine of theta is the y coordinate, and tangent of theta is the
1:54:11ratio. This video gives three properties of the trig function sine and cosine, that can
1:54:16be deduced from the unit circle definition. Recall that the unit circle definition of
1:54:22sine and cosine for an angle theta is that cosine theta is the x coordinate. And sine
1:54:32of theta is the y coordinate for the point on the unit circle at angle theta. The first
1:54:41property is what I call the periodic property. This says that the values of cosine and sine
1:54:48are periodic with period two pi. And what that means is that if you take cosine of an
1:54:54angle plus two pi, you get the same thing as just if you took cosine of the angle So,
1:55:00when we write this down, we're assuming that theta is measured in radians. If we want to
1:55:08measure theta in degrees, the similar statement is that cosine of theta plus 360 degrees is
1:55:15equal to cosine of theta, we can make the same statements for sine, sine of an angle
1:55:22plus two pi is equal to sine of the original angle. Here, the angle being measured in radians.
1:55:30If we want to measure the angle in degrees, the statement is that sine of theta plus 360
1:55:36is equal to sine of theta, we can see why this is true from the unit circle definition
1:55:43of sine and cosine. This is our angle theta, then theta plus two pi, the plus two pi adds
1:55:53a full turn around the unit circle to our angles, so we end up at the same place, theta
1:55:59and theta plus two pi are just two different names for the same location on the unit circle.
1:56:07And since sine and cosine give you the y and x coordinates of that point on the circle,
1:56:11they have to have the same value. Similarly, if we consider an angle theta, and an angle
1:56:18theta minus two pi, the minus two pi means we go the other direction around the unit
1:56:25circle clockwise, we still end up in the same place. And therefore, cosine of theta minus
1:56:32two pi, the x coordinate of that position is the same thing as cosine of theta, sine
1:56:38of theta minus two pi is the same thing as sine of theta, the same statements hold if
1:56:44we add or subtract multiples of two pi. For example, cosine of theta plus four pi is still
1:56:51the same thing as cosine of theta, this time, we've just gone to turns around the unit circle
1:56:57and still gotten back to the same place. So if we want to find cosine of five Pi, that's
1:57:04the same thing as cosine of pi plus four pi, which is the same thing as cosine of pi. Thinking
1:57:13about the unit circle, pi is halfway around the unit circle. So cosine of pi means the
1:57:20x coordinate of this point right here. Well, that point has coordinates negative one, zero,
1:57:26so cosine of pi must be negative one. If I want to take sine of negative 420 degrees,
1:57:33well, that's sine of negative 360 degrees, minus 60 degrees, which is the same thing
1:57:42as sine of minus 60 degrees. Thinking about the unit circle, minus 60 degrees, means I
1:57:50start at the positive x axis and go clockwise by 60 degrees that lands me about right here.
1:57:58And so that's one of the special angles that has an x coordinate of one half a y coordinate
1:58:05of negative root three over two. And therefore sine of negative 60 is negative root three
1:58:12over two the y coordinate. The next property I call the even odd property, it says that
1:58:19cosine is an even function, which means that cosine of negative theta is the same thing
1:58:25as cosine of theta, while sine is an odd function, which means that sine of negative theta is
1:58:33the negative of sine of theta. To see why this is true, let's look at an angle theta.
1:58:42And the angle negative theta. A negative angle means you go in the clockwise instead of counterclockwise
1:58:48direction from the positive x axis. The coordinates of this point by definition, r cosine theta
1:58:57sine theta, whereas the coordinates of this point are cosine of negative theta, sine of
1:59:04negative theta. But by symmetry, these two points have the exact same x coordinate, and
1:59:13therefore cosine of theta must equal cosine of negative theta, while their y coordinates
1:59:20have the same magnitude, but opposite signs. This one's positive and this one's negative.
1:59:27Therefore, sine of negative theta is the negative of sine of theta. Let's figure out if tan
1:59:34of theta isn't even or odd function. Well, we know that 10 of negative theta tangent
1:59:42by definition, is sine over cosine. Well, we know that sine of negative theta is the
1:59:51negative of sine of theta, whereas cosine of negative theta is cosine of theta. Therefore
2:00:00We're getting negative sine theta over cosine theta, which is negative tan of theta. Since
2:00:09tan of negative theta is the negative of tan of theta, tan theta is an odd function. The
2:00:20last property on this video is the Pythagorean property, which says that cosine of theta
2:00:25squared plus sine of theta squared is equal to one. A lot of times this property is written
2:00:32with this shorthand notation, cosine squared theta plus sine squared theta equals one.
2:00:41But this notation, cosine squared theta just means you take cosine of theta and square
2:00:48it. This property is called the Pythagorean property, because it comes from the Pythagorean
2:00:55Theorem. Let me draw a right triangle on the unit circle. I'll call this angle theta. So
2:01:02the coordinates of this endpoint here are cosine theta sine theta. Since this is supposed
2:01:08to be a unit circle, the hypothesis of my right triangle has length one, the base of
2:01:15my right triangle is just cosine theta, same thing as the x coordinate of this point. And
2:01:22the height of my triangle is the y coordinate of the point sine theta. Now the Pythagorean
2:01:28theorem says that this side length squared plus that side of that squared equals one
2:01:35squared. Since one squared is the same thing as one, that gives me the Pythagorean property.
2:01:42But the green property is handy for computing values of cosine given values of sine and
2:01:48vice versa. And this problem, we're told that sine of t is negative two sevenths. And T
2:01:54is an angle that lies in quadrant three. When we say the angle lies in quadrant three, that
2:02:01means the terminal side of the angle lies here in quadrant three. One way to find cosine
2:02:08of t is to use the fact that cosine squared t plus sine squared t is equal to one. That
2:02:18is cosine of t squared plus negative 2/7 squared is equal to one, I can write this as cosine
2:02:29of t squared plus 4/49 is equal to one. And so cosine of t squared is equal to one minus
2:02:394/49, which is 4540 nights. taking the square root of both sides, that goes for the cosine
2:02:48t is plus or minus the square root of 45 over 49, that's plus or minus the square root of
2:02:5645 over seven. Now, since we're in the third quadrant, we know that cosine of t, which
2:03:02represents the x coordinate of this point, must also be negative. Therefore, cosine of
2:03:09t is going to be negative square root of 45 over seven. It's also possible to solve this
2:03:17problem using the Pythagorean theorem for right triangles directly. If we look at the
2:03:23fact that sine of t is negative two sevens and ignore the negative sign for now, we can
2:03:28think of this information as telling us that we have a right triangle, angle theta, whose
2:03:35opposite side is to and whose high partners is seven. We call this side here a them but
2:03:43Tiger in theorem says us a squared plus two squared is seven squared. So a squared plus
2:03:50four is 49. So a squared is 45. And a is plus or minus the square root of 45. Since I'm
2:03:58worrying about a triangle, I'm going to use the positive value. Now, cosine of t is going
2:04:07to be adjacent over hypotenuse. So that's going to be the square root of 45 over seven.
2:04:16Now I go back to thinking about positive and negative signs. And I noticed that since I'm
2:04:22in the third quadrant, my co sign needs to be negative, so I just stick a negative sign
2:04:28in front. This alternative solution uses many of the same ideas as the previous solution,
2:04:35and ultimately gets us the same answer. This video gives three properties of trig functions,
2:04:42the periodic property, the even odd property, and the Pythagorean property. This video is
2:04:51about the graphs of sine and cosine. I want to graph the functions y equals cosine t and
2:04:57y equals sine t, where t is in radians. I'm going to think of this being the t axis, and
2:05:03this being the y axis. One way to do this is to plot points. So I'll fill in this chart,
2:05:10using my knowledge of special angles on the unit circle. These points will be easier to
2:05:16graph, if I convert them all to decimals. Now a plot the points for cosine and connect
2:05:22the dots to get a graph of y equals cosine t from t equals zero to t equals two pi. To
2:05:30continue the graph for t values less than zero or bigger than two pi, I could plot more
2:05:36points. Or I could just use the fact that the cosine values repeat. If I add or subtract
2:05:43two pi to the my angle T, I'll be at the same place on the unit circle. So my cosine will
2:05:50be exactly the same. Therefore, my values of cosine, which are represented by my y values
2:05:56on this graph, repeat themselves. For example, when my T value is two pi plus pi over six
2:06:03about like here, it's cosine is the same as the cosine of just pi over six. So I'll take
2:06:09this dot here and repeat it over here. Similarly, the when t is like two pi plus pi over four,
2:06:18I get the same value of cosine is when it's just pi over four. So this.is going to repeat,
2:06:24and I can continue repeating all my dots. This one repeats over here at two pi plus,
2:06:31say pi over three. And so my whole graph will repeat something like this. It also repeats
2:06:39on this side, something like this. Since subtracting two pi from my t values will also give me
2:06:48the same value of cosine. We can also plot points to get a graph for sine and extend
2:06:55it by repetition. Going forward, I'll usually write the function sine and cosine as y equals
2:07:03cosine of x and y equals sine of x. When I read it this way, notice that x now refers
2:07:11to an angle, while y refers to a value of cosine, or sine. That's a different meaning
2:07:20of x and y, compared to when we're talking about the unit circle, where x refers to the
2:07:24cosine value, and y refers to the sine value. Now let's look at some properties of the graphs
2:07:30of sine and cosine. The first thing you might notice is that the graph of cosine and the
2:07:36graph of sine are super similar to each other. In fact, you can think of the graph of cosine
2:07:42as just being the graph of sine shifted to the left by pi over two. So we can write cosine
2:07:50of x as the sine function of x plus pi over two, since adding pi over two on the inside,
2:08:00move the graph horizontally to the left by pi over two. Or we can think of the graph
2:08:07of sine as being constructed from the graph of cosine by shifting the cosine graph right
2:08:13by pi over two, that means we can write sine of x as equal to cosine of x minus pi over
2:08:19two, since subtracting pi over two on the inside, shifts the cosine graph to the right
2:08:25by pi over two. Next, let's look at domain and range. The domain of sine and cosine is
2:08:33all real numbers. Alright, that is negative infinity to infinity, but the range is just
2:08:38from negative one to one. That makes sense, because sine and cosine come from the unit
2:08:44circle. The input values for the domain come from angles. And you can use any numbers and
2:08:50angle positive negative as big as you want, just by wrapping a lot of times around the
2:08:54circle, the output values for the range, that is the actual values of sine and cosine come
2:09:00from the coordinates on the unit circle. And those coordinates can't be any bigger than
2:09:05one or any smaller than negative one. So that gives us a range. As far as even an odd behavior,
2:09:11you can tell from the graph, here's cosine, that it's symmetric with respect to the y
2:09:16axis, and so it must be even. Whereas the graph of sine is symmetric with respect to
2:09:21the origin and must be odd. The absolute maximum values of these two functions is one and the
2:09:28absolute minimum value is negative one. We can also use the words midline amplitude and
2:09:36period to describe these two functions. The midline is the horizontal line, halfway in
2:09:43between the maximum and minimum points. Here the mid line is y equals zero. The amplitude
2:09:52is the vertical distance between a maximum point and the midline. You can also think
2:09:58of the amplitude as the vertical distance between a minimum point and the midline, or
2:10:03as half the vertical distance between a min point and a max point. For the cosine function
2:10:09and the sine function, the amplitude is one. A periodic function is a function that repeats
2:10:16at regular horizontal intervals, the horizontal length of the smallest repeating unit is called
2:10:22the period. For y equals cosine of x, the period is two pi. Notice that the period is
2:10:31the horizontal distance between successive peaks, or maximum points, or between successive
2:10:38troughs, or minimum points. algebraically, we can write cosine of x plus two pi equals
2:10:46cosine of x and sine of x plus two pi equals sine of x to indicate that the functions repeat
2:10:55themselves over an interval of two pi and have a period of two pi. In this video, we
2:11:02graphed y equals cosine of x and y equals sine of x. and observe that they both have
2:11:11a midline at y equals zero, an amplitude of one and a period of two pi. Sine you sort
2:11:23of functions are functions that are related to sine and cosine by transformations like
2:11:29stretching and shrinking and shifting. This video is about graphing these functions. Let's
2:11:36start by graphing the function y equals three sine of 2x. This function is related to the
2:11:43function y equals sine x. So I'll graph that first. Now, the three on the outside stretches
2:11:50this graph vertically by a factor of three, while the two on the inside compresses that
2:11:58horizontally by a factor of one half. If instead I want to graph y equals three sine 2x plus
2:12:05one, this plus one on the outside shifts everything up by one unit. Let's compare the midline
2:12:14amplitude and period of our original y equals sine x are transformed y equals three sine
2:12:252x, and are further transformed y equals three sine 2x plus one, the original sign has a
2:12:32midline at y equals zero, an amplitude of one and a period of two pi. For the transformed
2:12:42function, y equals three times sine of 2x. The two on the inside shrinks everything horizontally
2:12:49by a factor of one half. So it changes the period of two pi into a period of one half
2:12:58times two pi, which just pi. Since the two on the inside only affects x values and horizontal
2:13:08distances, it doesn't affect the midline, which is a y value, or the amplitude, which
2:13:13is a vertical distance. But the three on the outside does affect these things. Well, in
2:13:18particular, it affects the amplitude, since everything is stretched out vertically by
2:13:23a factor of three, the amplitude of one get stretch to an amplitude of three. In this
2:13:31case, the midline doesn't actually change, because multiplying a y value of zero by three
2:13:37is still a y value of zero. Now on the third function, we've taken the second function
2:13:45and added one on the outside, so that shifts everything up by one. Therefore, instead of
2:13:50having a midline at y equals zero, we now have a midline at y equals one, the amplitude
2:13:57doesn't change though, it's still three, because shifting everything up by one doesn't affect
2:14:01the distance between the mid mine and the end the maximum point. Also, the period is
2:14:07still pi since the period is a horizontal measure, and adding one on the outside only
2:14:12affects vertical things. Next, let's graph the function y equals three times sine of
2:14:18two times quantity x minus pi over four. This function is very closely related to the first
2:14:25function we graphed on the previous page, that was y equals three sine of 2x. In fact,
2:14:32if we give the name f of x to that function, and maybe we can call g of x, this other function,
2:14:42then we can get g of x by taking f of x and plugging in x minus pi is pi over four in
2:14:50for x. In other words, g of x is f of x minus pi over four. This relationship gives mean
2:15:00idea for graphing g of x the function we want to graph, we can first graph f of x, we already
2:15:06did that on the previous page. And then we can shift its graph to the right by pi over
2:15:16four, because that's what you do when you subtract a number on the inside of a function.
2:15:23So here's the graph of y equals three sine 2x. Recall that it's just the graph of sine
2:15:31stretched vertically by a factor of three, and shrunk horizontally by a factor of one
2:15:35half. Now, to graph the function that I want, I'm going to shift this graph over by pi over
2:15:42four to the right. Notice that since I had my function written in factored form, I could
2:15:48just read off the horizontal shift. But if I had written it instead, as y equals three
2:15:57sine 2x minus pi over two, which is algebraically equivalent, it would be easy to get confused
2:16:06and think that I needed to shift over by pi over two. So it's best to factor first, before
2:16:13figuring out what the shift is, we're factoring out the coefficient of x. If instead, we wanted
2:16:20to graph this function, same as the one we just graphed, it's just with a minus one on
2:16:25the outside, that minus one would just bring everything down by one. Let's take a moment
2:16:30to look at midline amplitude, and period for the original parent function, y equals sine
2:16:41of x, and our final transformed function, y equals three sine of two times quantity
2:16:48x minus pi over four minus one, our original sine function has midline at y equals zero
2:16:56amplitude of one and period of two pi. For our transform function, the three on the outside
2:17:04stretches vertically, so it makes the amplitude three, the minus one on the outside shifts
2:17:14everything down by one. So it brings the midline, y equals zero, down to Y equals negative one,
2:17:24the two on the inside, shrinks everything horizontally by a factor of one half. So the
2:17:30period becomes one half times two pi, which is pi. Finally, there's a horizontal shift
2:17:39going on, are transferring function shifts to the right, by pi over four, this horizontal
2:17:51shift is sometimes called the phase shift. The function we just analyzed was y equals
2:18:00three sine 2x minus pi over four minus one, which could also be written as y equals three
2:18:10sine 2x minus pi over two minus one. This is a function of the form y equals a sine
2:18:19B x minus c plus d, where b is positive. If we have a function of this form, or the similar
2:18:26function with cosine in it, then we know that the midline is going to be at y equals D.
2:18:35That's because the original midline of sine or cosine at y equals zero gets shifted up
2:18:42by D, we know that the amplitude is going to be a because this A multiplied on the outside
2:18:51stretches everything vertically by a factor of a to be a little more accurate, we should
2:18:57say the amplitude is the absolute value of A in case a is negative. If a is negative,
2:19:02then that amounts to a vertical reflection, or a reflection over the x axis. We know that
2:19:10the period of the original sine or cosine is two pi. And we know that this factor of
2:19:16B amounts to a horizontal shrink by a factor of one over B, or I guess it could be a horizontal
2:19:23stretch by a factor of one over b if b is less than one. So because we're starting with
2:19:30a period of two pi, and we're multiplying by one over B, our new period is going to
2:19:37be two Pi over B. The trickiest thing is the horizontal shift. And to get that right, I
2:19:46like to factor out this be from my equation. So instead of writing y equals a cosine bx
2:19:55minus c plus d, I'm going to write y equals A cosine B times quantity x minus c over b
2:20:03plus d. Similarly, if it's a sine function, I write y equals a sine B times x minus quantity
2:20:10c over b plus d, then I can read off the horizontal shift as C over B. And that'll be a shift
2:20:19to the right, if C ever be as positive and a shift to the left, if C over b is negative,
2:20:25this might seem backwards from what you're used to. But it's because we have that minus
2:20:28sign there. So if C over B is positive, we're actually subtracting on the inside. So that
2:20:32shifts right, if C of b over b is negative minus a negative is actually adding something,
2:20:37and that's why it shifts it to the left. So as one final example, say I wanted to graph
2:20:42y equals 1/3, cosine of one half x plus three minus five, that would have a midline at y
2:20:57equals minus five, an amplitude of 1/3, a period of two pi divided by one half, which
2:21:09is four pi, and a horizontal shift. But a rewrite this horizontal shift of six units
2:21:22to the left, the horizontal shift is sometimes called the phase shift. And that's all for
2:21:33graphs of sinusoidal functions. This video is about graphing the trig functions, tangent,
2:21:39secant, cotangent and cosecant. To gain an intuition for the graph of y equals tangent
2:21:47of x, I think it's handy to look at the slope of a line at angle theta on the unit circle.
2:21:56The slope of this line is the rise over the run. But the rise is given by sine of theta,
2:22:06and the run is given by cosine of theta. So the slope is given by sine theta over cosine
2:22:16theta, which is simply tan of theta. So if I want to graph y equals tan of x, I can think
2:22:24of x as being the angle and y as being the slope of the line at that angle. Notice if
2:22:36the angle is zero, the slope is zero. But as the angle increases towards pi over two,
2:22:48the slope gets bigger and bigger heading towards infinity. As the angle goes from zero towards
2:22:59negative pi over two, the slope is getting negative and heading towards negative infinity,
2:23:06at exactly pi over two and negative pi over two, we have a vertical line. And so the slope
2:23:15is undefined. Using this information, let's graph a rough sketch of y equals tan x. Remember,
2:23:23we're thinking of x as the angle and y as the slope, we're going to go between an angle
2:23:31of negative pi over two and pi over two. So we said that the slope was zero when the angles
2:23:39zero, and then it heads up towards positive infinity, as we go towards the angle goes
2:23:46towards pi over two with an undefined value at pi over two, it goes negative heading towards
2:23:52negative infinity as the angle heads towards negative pi over two, also, with an undefined
2:23:58value at negative pi over two, you can also verify that for angles slightly bigger than
2:24:04pi over two, we have the same line as for angles that are approaching negative pi over
2:24:12two, and therefore this picture repeats. And it turns out that tangent is periodic with
2:24:26period not to pi, like sine and cosine, but just pi. The period of pi makes sense because
2:24:35if you take a line and rotate it by 180 degrees, it's the same line with the same slope and
2:24:43therefore has the same value of tangent in this graph of y equals 10x knows that the
2:24:51x intercepts all right values of x of the form negative two pi negative pi Zero, pi,
2:25:02two pi, etc, you can write that as pi times k, where k is an integer that is a positive
2:25:13or negative whole number or zero. This makes a lot of sense because tangent of x is sine
2:25:21of x over cosine of x. And so you're gonna get x intercepts, that's where y is zero,
2:25:28which is where the numerator is zero, and sine x is zero, at values of the form pi,
2:25:37two pi, and so on. From the graph, you can see the vertical asymptotes are at values
2:25:44like negative three pi over two, negative pi over two, pi over two, and three pi over
2:25:51two, these values can be written as pi over two times k, where k is an odd integer. Again,
2:26:00this makes sense from the definition of tangent, since the vertical asymptotes will occur where
2:26:05the denominator is zero, and cosine x is zero, at numbers, like negative pi over two pi over
2:26:12two, three pi over two, and so on, the domain of tangent is the x axis for which it's defined.
2:26:20So that's going to be everything except for the vertical asymptotes, we can write that
2:26:24as x such that x is not equal to pi over two times k, for K, an odd integer. The range
2:26:35or the y values go all the way from negative infinity to infinity. And the period, as we
2:26:41mentioned previously, is pi. Since the smallest repeating unit has a horizontal width of pi,
2:26:53to graph y equals secant x, I'm going to remember that secant is one over cosine. So if I start
2:27:01with a graph of cosine, I can take the reciprocal of the y values to get the graph of secant,
2:27:08the reciprocal of one is one, the reciprocal of zero is undefined, so I'm not going to
2:27:15have a value at pi over two, negative pi over two, three pi over two, or negative three
2:27:23pi over two. When I take the reciprocal of numbers, just less than one, I'm going to
2:27:29get numbers just greater than one, but I would take the reciprocal of positive numbers getting
2:27:34close to zero, I'm going to get really big positive numbers going up towards infinity.
2:27:41Similarly, on the other side, over here, I have numbers close to zero, but negative,
2:27:49so their reciprocals will be negative numbers heading towards negative infinity, the reciprocal
2:27:55of negative one is negative one. And similarly here, so I'm getting kind of positive and
2:28:02negative buckets and upside down buckets as the graph of my sequence. Notice that secant
2:28:11has a period of two pi, which makes sense, since cosine has a period of two pi, it has
2:28:17a range that goes from negative infinity to negative one inclusive, and from one to infinity.
2:28:25That makes sense because the range of cosine is between one and negative one, and we're
2:28:30taking the reciprocal of those values. The domain is everything except for the vertical
2:28:35asymptotes. Now the vertical asymptotes are where cosine is zero. So that is at values
2:28:48of the form pi over two, three pi over two, etc. That's values of the form pi over 2k,
2:28:54where k is an odd integer. So the domain is going to be x values such that x is not equal
2:29:03to pi over 2k. For K and odd integer. The x intercepts of secant Well, it doesn't have
2:29:13any, because you can't take one over something and get the numbers zero for your y value.
2:29:23We've seen the graph of y equals tan x and y equals secant x. This is the graph of y
2:29:32equals cotangent x. It looks similar to the graph of tangent x, it's just a decreasing
2:29:38function instead of an increasing one, and it has its vertical asymptotes and its x intercepts
2:29:45in different places. Finally, this green graph is the graph of y equals cosecant x. It's
2:29:55related to the graph of sine x, since cosecant is one over sine x. And in fact, if I draw
2:30:02the graph of sine x in between, you can see how it kind of bounces off. Because it's the
2:30:09reciprocal. I encourage you to memorize the general shape of these graphs, you can always
2:30:15figure out the details by thinking how about how they're related to the graphs of cosine
2:30:24of x, and sine x. This video is about the graphs of transformations of tangent, secant
2:30:33cotangent and cosecant. Before I graph this fancy function, I'm going to graph the related
2:30:40parent function, y equals cosecant of x. Recall that cosecant of x is one over sine of x.
2:30:50So the graph of cosecant of x has a vertical asymptotes, where sine of x is zero, that
2:30:57puts the vertical asymptotes at pi, two pi, zero, and so on. Because seeing n of x is
2:31:07one where sine is one. So here, and so it looks something like this. Now, when I graph
2:31:16my fancy function, I know that the two on the outside does a vertical stretch by two,
2:31:24the plus one shifts up by one, the PI means everything shrinks horizontally by one over
2:31:38pi. And so the period, instead of being two pi, like it is for cosecant, it's going to
2:31:44be two pi divided by pi, which is two. Finally, this pi over two has to do with a horizontal
2:31:53shift. But in order to see by how much we horizontal shift, we need to factor out the
2:31:58pie first. So if I rewrite my function, as two times cosecant pi times x plus a half,
2:32:08plus one, I can see the horizontal shift is one half to the left. Let me do those transformations
2:32:15in pieces. First, I'll do the vertical motions. So I'll leave the horizontal axis as is I'll
2:32:22stretch by a factor of two. And now I'll shift up by one. Next, I'll worry about the horizontal
2:32:30motion, instead of a period of two pi, I'm going to have a period of two. But in addition,
2:32:38everything is going to be shifted left by one half. So instead of having my vertical
2:32:43asymptotes at zero, pi, two pi and so on, they would be at 012, and so on, except everything
2:32:57shifted over by one half. So that means the vertical asymptotes will actually be at negative
2:33:05a half, a half, one and a half, and so on. Another way to find the vertical asymptotes
2:33:16is by noting that cosecant is one over sine, so therefore, cosecant pi x plus one half
2:33:27does not exist when a sine pi x plus one half equals zero, ie when pi times x plus one half
2:33:44is equal to a multiple of pi. canceling the PI's and solving for x, that's when x is some
2:33:54number minus a half. So those are exactly the values of say, one half, three half, five
2:34:03halves, and so on that we've drawn. So once we've got the vertical asymptotes in place,
2:34:09we can see that our graph fits in between them. This purple piece got shrunk horizontally
2:34:17and shifted by a half, so it ends up hugging the x axis. I'll draw it in right here. And
2:34:24all the other pieces also get shrunk horizontally and shifted. So my final graph looks something
2:34:32like this. In this next problem, we're given the graph and we need to find the equation,
2:34:41the shape of the graph. And the fact that it's increasing tells us it must be a tangent
2:34:49graph. But y equals tan x by itself would have a period of pi. Let's check what our
2:34:58graph has a period of need to find the horizontal width of one complete cycle, it's probably
2:35:06easiest to measure if I sketch in the vertical asymptotes, which seem to run from an x value
2:35:12of negative three to an x value of one. So that's a horizontal distance of four. So our
2:35:18graph might be something more like tan of Bx, where pi over B has to be four. That means
2:35:30that pi equals four B. And so B is pi over four. So our graph is probably more closely
2:35:38related to Y equals tan of pi over 4x. But if it were exactly y equals tan of pi over
2:35:474x, then that middle point right here would be at the origin. And instead, that middle
2:35:54point is at negative one, one, that means my tangent graph is shifted left by one, and
2:36:03up by one, the left by one, I can accomplish by writing x plus one, when I have the Pi
2:36:14over four factored out the up by one I get by writing a plus one on the outside. Now
2:36:21as possible, there might also be a vertical stretch factor out front, an easy way to check
2:36:27for that is by plugging in x equals zero and seeing whether what kind of number out in
2:36:33front I might need to get this y intercept of two. So I'm going to write y equals a tan
2:36:41pi over four quantity x plus one plus one, plug in point 02. So that's two equals a tan
2:36:54pi over four, zero plus one plus one. And I can solve for A since tan of pi over four
2:37:03is just one, this equation simplifies to two equals a times one plus one, which means that
2:37:11a has to be one. So in fact, this equation here is our final answer for this graph. To
2:37:20summarize, if we have a function of the form y equals a tan bx minus c plus d, the number
2:37:27a accomplishes a vertical stretch. The number B changes the period from pi to pi over B,
2:37:44the number D shifts vertically by D. And finally, to figure out what the number c is, we have
2:37:57to rewrite our equation in factored form factoring out the B. And so we see that there's a phase
2:38:06shift right by C over B. The same things hold. If instead of tangent, we have a graph of
2:38:18cotangent, we can say the same things for our graph of secant and a graph of cosecant.
2:38:26The only thing to be careful of is that B now changes the period from two pi to two
2:38:36pi over B, simply because the original period of secant and cosecant is two pi, whereas
2:38:47the original period of tan and co tan is just pi. If you know what the original graphs of
2:38:57tan, secant cotangent and cosecant look like, you can use your knowledge of transformations
2:39:03to graph more complicated functions. This video defines the standard inverse trig functions,
2:39:11sine inverse cosine inverse and tan inverse. In this crazy looking graph, please focus
2:39:21first on the thin black line. This is a graph of y equals sine x. The graph of the inverse
2:39:27of a function can be found by flipping the graph of the original function over the line
2:39:32y equals x. I've drawn the flipped graph with this blue dotted line. But you'll notice that
2:39:40the blue dotted line is not the graph of a function because it violates the vertical
2:39:44line test. So in order to get a function, that's the inverse of y equals sine x, we
2:39:51need to restrict the domain of sine of x. We'll restrict it to this piece that's drawn
2:39:58with a thick black line. If I invert that piece, by flipping it over the line y equals
2:40:04x, I get the piece drawn with a red dotted line here. And that piece does satisfy the
2:40:10vertical line test. So it is in fact, a function of the regular sine x has domain from negative
2:40:17infinity to infinity are restricted sine x has domain from negative pi over two to pi
2:40:24over two. It's, its range is still from negative one to one, just like the regular sine x.
2:40:33Because I've taken the biggest possible piece of the graph, whose flipped version is still
2:40:39a function. The inverse sine function is often written as arc sine of x. And since inverting
2:40:49a function reverses the roles of y at x, it reverses the domain and the range. So arc
2:40:57sine of x, the inverse function has domain from negative one to one, and range from negative
2:41:03pi over two to pi over two, which seems plausible from the graph. Now an inverse function undoes
2:41:11the work a function. So if the function sine takes angles theta, two numbers, x, then the
2:41:22inverse sine, or arc sine takes numbers x, two angles theta. For example, since sine
2:41:33of pi over two is one, arc sine of one is pi over two. And in general, the output of
2:41:42arc sine of x is the angle between negative pi over two and pi over two whose side is
2:41:51x. y is equal to arc sine x means that x is equal to sine of y. But since there are many
2:41:59angles, y who sine is x, right, they all differ by multiples of two pi, we specify also, that
2:42:06y is between negative pi over two and pi over two. That was the whole point of doing this
2:42:13domain restriction in order to get a well defined inverse value. There's an alternate
2:42:19notation for inverse sine. Sometimes it's written as sine to the negative one of x.
2:42:25But this notation can be confusing, so be careful. In particular, sign to the negative
2:42:33one of x does not equal one over sine of x. One over sine of x the reciprocal function
2:42:41is another word for cosecant of x. But sine to the negative one of x is another word for
2:42:48arc sine of x, the inverse sine function, which is not the same thing as the reciprocal
2:42:55function, let's go through the same process to build an inverse cosine function, we start
2:43:03with a graph of cosine of x, we flip it over the line y equals x to get the blue dotted
2:43:10line. But the blue dotted line is not a function. So we go back and restrict the domain for
2:43:16our original cosine of x to just be between zero and pi. The resulting red graph now satisfies
2:43:24the vertical line test. So it's a proper inverse function. Our restricted cosine has domain
2:43:30from zero to pi, and range from negative one to one. And so our inverse function, arc cosine
2:43:39has domain from negative one to one, and range from zero to pi. Since cosine takes us from
2:43:47angles to numbers, arc cosine takes us from numbers back to angles. For example, cosine
2:43:58of pi over four is the square root of two over two. So arc cosine of the square root
2:44:04of two over two is equal to pi over four. arc cosine of x is the angle between zero
2:44:14and pi, whose cosine is x. In other words, y equals r cosine of x means that x is equal
2:44:24to cosine of y. And y is between zero and pi, since otherwise, there'd be lots of possible
2:44:32answers for an angle y whose cosine is x. The alternative notation for arc cosine is
2:44:40cosine inverse. And again, we have to be careful. cosine to the negative 1x is not the same
2:44:49thing as one over cosine of x. one over cosine of x is also called secant of x. cosine to
2:44:57the negative 1x means arc cosine On the inverse function, and these two things are not the
2:45:03same. Finally, let's look take a look at inverse tangent function. Here's a graph of tangent,
2:45:14and black. These vertical lines aren't really part of the function there, just vertical
2:45:19asymptotes. So in order to get the actual function, when we flip over the line y equals
2:45:26x, we take just one piece of the tangent function. Here, we've taken the piece marked in black,
2:45:34we flip that over the line y equals x, we get this piece in red, which is actually a
2:45:40function because it satisfies the vertical line test. Now, you might ask, would it be
2:45:45possible to pick a different piece of the tangent function to invert? And the answer
2:45:52is, yes, we could do that. And on another planet, maybe mathematicians do that. But
2:45:58on our planet, we use the convention that we pick this piece of tangent to invert, which
2:46:05is kind of a convenient choice since it's centered here around the origin. In the previous
2:46:11two examples, our choice of restricted domain for sine and for cosine was also a convention
2:46:17that led to a conveniently defined inverse function. In any case, based on our choice,
2:46:24our restricted tan function has domain from negative pi over two to pi over two. We don't
2:46:31include the endpoints in that interval, because our restricted tan function has vertical asymptotes
2:46:37at negative pi over two and pi over two, so it's not defined there. The range of our restricted
2:46:43tan function is from negative infinity to infinity. Therefore, arc tan of X has domain
2:46:53from negative infinity to infinity, and range from negative pi over two to pi over two.
2:47:00Once again, tangent is taking us from angles to numbers. So arc tan is taking us from numbers
2:47:08to angles. For example, tangent of pi over four is one, and therefore, arc tan of one
2:47:20is pi over four. So arc tan of x means the angle between negative pi over two and pi
2:47:30over two whose tangent is x. y is equal to arc tan x means that x is equal to tangent
2:47:40of y. And that y is between negative pi over two and pi over two. The inverse tan function
2:47:52can also be written as 10 to the minus one of x. And once again, tan inverse of x means
2:48:02the inverse trig function, arc tan of x. And it's not equal to one over tan of x, which
2:48:15is called cotangent of x. And that's all for this video on the three basic inverse trig
2:48:25functions. sine inverse x, also known as arc sine of x, cosine inverse x, also known as
2:48:33arc cosine x, and tan inverse x, also known as arc tan x. This video is an introduction
2:48:41to solving trig equations. Let's start with the equation two cosine x plus one equals
2:48:48zero, I want to find all the solutions in the interval from zero to two pi, and then
2:48:53get a general formula for all solutions, not just those in that interval. Let me start
2:48:59by rewriting this equation to isolate the tricky part, which is cosine of x. So I'm
2:49:06going to write two cosine x equals negative one, and then divide both sides by two. Now
2:49:15I'm looking for the angles x between zero and two pi whose cosine is negative one half.
2:49:24Since negative one half is one of the special values on the unit circle, I can use my knowledge
2:49:30of the unit circle to see that the angle between zero and two pi must be either two pi over
2:49:39three, or four pi over three. My answer needs to include both of these values. There are
2:49:48no other spots on the unit circle whose cosine is negative one half. But there are more angles
2:49:54because we can always take one of these angles and add multiples of two pi to it. So if I
2:49:59want to Find all solutions, I can take these two principles solutions, two pi over three,
2:50:07and four pi over three, and simply add multiples of two pi to them, for example, two pi over
2:50:13three plus two pi, or two pi over three minus two pi, two pi over three plus four pi, and
2:50:21so on. A much more efficient way to write this is to write two pi over three plus two
2:50:30pi times k, any integer that is any positive or negative whole number, or zero. Similarly,
2:50:41I can write four pi over three plus two pi k, to capture all solutions, based on the
2:50:50principal solution for pi over three by adding and subtracting multiples of two pi. This
2:50:55is my final solution. Next, let's look at a trig equation involving tangent. As usual,
2:51:05I'm going to start out by cleaning things up and isolating the tricky part, which in
2:51:09this case is tangent. So let me add tangent to both sides. That'll give me three, tan
2:51:19x equals the square root of three. And so tan x is the square root of three over three,
2:51:27the square root of three over three looks suspiciously similar to value the value of
2:51:32square to three over two, which is a special value on my unit circle. So my suspicion is
2:51:37that my unit circle will again, help me find this value of x without a calculator. Recall
2:51:43that tan x is sine x over cosine x. So I'm looking for angles on the unit circle between
2:51:53zero and two pi with a ratio of sine over cosine will give me square root of three over
2:51:59three, I actually only need to look in the first quadrant, and the third quadrant, because
2:52:05those are the quadrants where a tangent is positive. And I really only need to look at
2:52:12angles whose either sine or cosine has a squared of three in it. So by trial and error, I can
2:52:19see that tan pi over six, which is sine pi over six over cosine pi over six, will give
2:52:32me one half over root three over two, that's the same thing as one half times two over
2:52:38three, which is one over root three. If I rationalize that, I get root three over three.
2:52:45So that value works. If I try tan of pi over three, instead, I get root three, which is
2:52:56not equal to root three over three. So pi over three doesn't work. Similarly, I can
2:53:02work out some the values in the third quadrant and see that seven pi over six works. But
2:53:11four pi over three does not. So my answer to part A includes just two values, pi over
2:53:21six, and seven pi over six. Now if I want to find all solutions, not just those in the
2:53:29interval from zero to two pi, I noticed that I can take one of these principal solutions
2:53:35and add multiples of two pi to it, because that'll give me the same angle. So I get pi
2:53:42over six plus two pi k, and pi over six, sorry, seven pi over six plus two pi k, any integer.
2:53:58This is a correct answer. But it's not as simple as it could be. Notice that seven pi
2:54:03over six over here on the unit circle is exactly pi more than pi over six. So instead of taking
2:54:10both of these and adding multiples of two pi to them, I could get all the same answers
2:54:15by just taking one of them and adding multiples of pi to it. So a more efficient answer is
2:54:22to say that x equals pi over six plus pi times k for K any integer. This will still capture
2:54:34all the same solutions. Because when k is even, I'll get this family of solutions. And
2:54:39when k is odd, I'll get this family. For example, when k is one pi over six plus one times pi
2:54:47is just the original seven pi over six. If you think about the fact that tangent has
2:54:53a period of pi instead of two pi, it makes a lot of sense that you should be able to
2:54:58write the solutions in the form. In this video, we solved basic trig equations by first, isolating
2:55:09sine, or tangent, or the same thing would work with cosine. And then using the unit
2:55:18circle to find principal solutions, Principal solutions are just solutions between zero
2:55:28and two pi. And then adding multiples of two pi to these principles solutions to get all
2:55:39solutions. For tangent, we noticed that it was equivalent to just use one principal solution
2:55:45and add multiples of pi instead of two pi. In a previous video, we looked at trig equations
2:55:53that can be solved using the unit circle. In this video, we'll solve trig equations
2:55:59that don't involve special values on the unit circle, and require a calculator instead to
2:56:04solve. Let's look at the equation two cosine t equals one minus cosine t. As usual, I'll
2:56:13start by simplifying and isolating cosine. Adding cosine to both sides, I get three cosine
2:56:19t equals one divided by three gives cosine t equals 1/3. Now 1/3 is not one of my special
2:56:28values on the unit circle, but I can locate it approximately cosine represents the x value
2:56:36of a point on the unit circle. So I'm looking for a point on the unit circle whose x coordinate
2:56:41is approximately 1/3. So maybe right here, and the symmetric angle right here, those
2:56:48are angles in the first and fourth quadrants. To get a decimal approximation for one of
2:56:54these angles, I can take arc cosine of both sides of my equation. This gives me t equals
2:57:02arc cosine of 1/3, which, using my calculator on radian mode, I get a decimal approximation
2:57:09of 1.2310. Up to four decimal places, it's important to use radian mode here. Since by
2:57:18convention, the answers to trig equations are expected to be in radians unless otherwise
2:57:25specified. It's also important to notice that these two equations are not precisely the
2:57:31same thing. Since cosine t equals 1/3, has infinitely many solutions if we don't restrict
2:57:40to a certain interval, and even if we restrict to the interval zero to two pi, it still has
2:57:45two different solutions. On the other hand, our T equals arc cosine of 1/3 has only one
2:57:53solution. And that solution lies between the angles of zero and pi. So this angle here,
2:58:04the 1.231, a radians must be giving us this angle, so there's got to be an angle between
2:58:10zero and pi. And to get the other angle this one, we can use the symmetry of the situation.
2:58:19This angle and this angle are the same size. So to get all the way around to here, we can
2:58:24go all the way to pi minus r 1.2312 radians. Using our calculator again, we can compute
2:58:33that two pi minus 1.2310 radians is equal to 5.05 to two radians. So our two solutions
2:58:44on our interval from zero to two pi, are given by 1.2310 and 5.05 to two radians. Let me
2:58:55emphasize that we found the second solution by taking two pi minus the first solution.
2:59:07Now it's straightforward to find all solutions, we can just take our two principal solutions
2:59:12and add multiples of two pi. It can be helpful to look at this solution graphically. To solve
2:59:21our equation, cosine t equals 1/3. We want to know where this graph of y equals cosine
2:59:31of t intersects the horizontal line at y equals 1/3. This intersection point here corresponds
2:59:40to our first solution of 1.2310 radiance. This intersection point here corresponds to
2:59:48our second solution. Notice that it occurs at a T value, that's two pi minus this T value
2:59:56here and all the other initial action points correspond to those first two intersection
3:00:03points shifted over by multiples of two pi. Next, let's look at the equation for sine
3:00:12x minus one equals two. I'll start by rewriting this as four sine x equals three, which is
3:00:20equivalent to sine x equals three fourths. Although three fourths is not a special value
3:00:26on the unit circle, we can still use the unit circle to get oriented and see how many solutions
3:00:32there should be in the interval from zero to two pi. Since sine represents the y coordinate
3:00:38of a point on the unit circle, we're looking for points on the unit circle whose y coordinate
3:00:42is three fourths. So that might be three fourths might be about here. So that would be an angle
3:00:47about right here. And about right here, so in the first and second quadrants, let's take
3:00:56the arc sine of both sides to find one of those angles. I'm getting one of the angles
3:01:03to be arc sine of three fourths, which my calculator says is 0.8481 radians, that must
3:01:12be this angle here. To get the other angle, I can use symmetry. Since this angle here,
3:01:22you can get by going up four pi minus my previous angle, pi minus 0.8481, is equal to 2.2935.
3:01:35So that's my second angle. Notice that this time, to get the second angle, we use pi minus
3:01:44the first angle, not to pi minus the first angle like we did for cosine. Once again,
3:01:51we can find all solutions simply by adding multiples of two pi, we can also locate the
3:01:57solutions on a graph of sine, we're looking for where the graph of y equals sine of x
3:02:05intersects with the horizontal line at y equals three fourths. And we can see that we have
3:02:13these two solutions where the second one, its x value is pi minus the x value this one,
3:02:23and then we have all these additional solutions that we can get by shifting over by positive
3:02:30and negative multiples of two pi from the first to finally, let's solve the equation
3:02:3510x equals four. If we simply take x to be tan inverse of four, which is 1.3258 radians,
3:02:48we'll get a single solution. But thinking about the graph of tangent, and intersecting
3:02:57that graph with the horizontal line y equals four, we see that there should actually be
3:03:05two solutions in the interval from zero to two pi. This makes sense, because the period
3:03:14of tangent is pi. So any one solution will get repeated. By intervals of pi. It also
3:03:24makes sense by thinking about the unit circle. Tangent, which can be thought of as the slope
3:03:31of a line at angle theta takes on all positive real values, both in the first quadrant and
3:03:38the third quadrant. So there are going to be two places on the unit circle where the
3:03:44tangent is for. We've already found this one. To find the other one, we just need to add
3:03:51pi. This gives us our answer for the interval zero to two pi. And we can add multiples of
3:03:58two pi to get all other solutions. Or if we want to be more efficient, we can just write
3:04:04the first solution plus multiples of pi instead of two pi. Because these two solutions differ
3:04:12by exactly pi. Taking the first solution, and then adding multiples of pi will capture
3:04:20all solutions corresponding to both of these principal solutions. To summarize, to find
3:04:28our solutions to cosine t equals m, we need to find first cosine inverse of m. Then take
3:04:38two pi a minus this solution and add multiples of two pi to both of those. To find solutions
3:04:47to the equation, sine t equals n. We can start by taking sine inverse of n using pi minus
3:04:57this and adding multiples of to pi. To find our solutions to tangent t equals P, we can
3:05:07take t equal tan inverse of P, and tan inverse of p plus pi and add multiples of two pi.
3:05:19Or we can just take 10 inverse of P, and add multiples of pi. Although the examples that
3:05:27we saw were more simple than some, because for the examples, we saw, the inverse trig
3:05:33function always landed us in the first quadrant, these methods will work more generally. That's
3:05:40all for this video on solving trig equations that require a calculator. This video is about
3:05:47identities involving trig functions like sine and cosine. But I want to start with some
3:05:52examples that just involve quadratic functions. If I want to find the solutions to this equation,
3:06:00I can rewrite it x squared minus 6x minus seven equals zero, factor it, x minus seven
3:06:09times x plus one equals zero, set the factors equal to 0x minus seven equals zero, or x
3:06:17plus one equals zero. And that gives me the solutions, x equals seven, or x equals negative
3:06:24one. Next, let's look at this more complicated equation. I'm going to try to solve that for
3:06:32x by multiplying out the right hand side. Next, our combined terms on the right hand
3:06:40side. So that gives me x squared minus 6x. On both sides, well, x squared minus 6x is
3:06:51equal to x squared minus 6x. That's true no matter what I plug in for x. And therefore,
3:06:58all values of x satisfy this equation, we can say that the solution set is all real
3:07:05numbers. The second equation is called an identity, because it holds for all values
3:07:16of the variable. The first equation, on the other hand is not an identity, because it
3:07:22only holds for some values of x and not all values. Please pause the video for a moment
3:07:28and try to decide which of the following three equations or identities that is, which of
3:07:34these equations hold for all values of the variable. To start out, you might want to
3:07:41test them by plugging in a few values of the variable and see if the equation holds. The
3:07:49first equation is not an identity. It does hold for some values of x. For example, if
3:07:56x equals zero, then sine of two times zero is zero, and two times sine of zero is also
3:08:05zero. So it does hold when x is zero. However, when x is say, pi over two, then sine of two
3:08:14times pi over two, that's the same thing as sine of pi, which is zero, but two times sine
3:08:21of pi over two, that's two times one, or two, and zero is not equal to two. So the equation
3:08:30does not hold for x equals pi over two. Since it doesn't hold for all values of the variable,
3:08:36it's not an identity. The second equation is an identity. You can build some evidence
3:08:44for this by plugging in numbers. For example, cosine of zero plus pi, which is negative
3:08:51one is the same thing as negative of cosine of zero. You can also check for example, that
3:08:58cosine of pi over six plus pi is the same thing as negative cosine of pi over six. But
3:09:04even if we check a zillion examples, that's just evidence, it's not a proof that the identity
3:09:09holds, we could have just gotten lucky with the values we picked, we can build a little
3:09:14bit stronger evidence by looking at graphs, I'm going to put theta on the x axis in a
3:09:22pie graph, y equals cosine of theta plus pi. That's just like the graph of cosine shifted
3:09:31over to the left by pi. On the other hand, if I graph y equals negative cosine theta,
3:09:38that's the graph of cosine theta, reflected across the x axis, which gives us the exact
3:09:46same graph. So graphing both sides gives us strong evidence that this equation is an identity
3:09:54it holds for all values of theta. Now the strongest evidence of all would be an algebraic
3:09:59proof, which we'll do later in the course, once we have a formula for the cosine of a
3:10:05sum of two angles. In the meantime, let's look at equation C. It turns out equation
3:10:13C is an identity. And we could build evidence for it again by plugging in values for x,
3:10:19or by graphing the left side and the right side separately, and checking to see that
3:10:25the graphs coincided. But for this example, I'm going to go ahead and do an algebraic
3:10:30verification. In particular, I'm going to start with the left side of the equation and
3:10:37rewrite things and rewrite things until I get to the right side of the equation. The
3:10:42first thing I'll rewrite is secant and tangent in terms of their constituent functions, sine
3:10:48and cosine. Since secant of x is one over cosine x, and tangent of x is sine x over
3:10:54cosine x, I can rewrite this expression as one over cosine x minus sine x times sine
3:11:05x over cosine x. I can clean up those fractions and write this as one over cosine x minus
3:11:15sine squared x over cosine x. Now, I noticed that I have two fractions with the same denominator.
3:11:23So I can pull them together as one minus sine squared x over cosine x. Next, I'm going to
3:11:33rewrite the numerator one minus sine squared x using the Pythagorean identity that says
3:11:40that cosine squared x plus sine squared x equals one, and therefore, one minus sine
3:11:49squared x is equal to cosine squared x just by subtracting sine squared x from both sides.
3:11:59So I can replace my numerator, one minus sine squared x with cosine squared x. And canceling
3:12:10one cosine from the top and from the bottom, that's the same thing as cosine of x, which
3:12:16is the right hand side that I was trying to get to. So a combination of a bunch of algebra,
3:12:22and the Pythagorean identity allows me to prove that this equation is true for all values
3:12:28of x, it's an identity. The best way to prove that an equation is an identity is to use
3:12:36algebra, and to use other identities, like the Pythagorean identity, to rewrite one side
3:12:46of the equation, till it looks like the other side. The best way to prove the net equation
3:12:52is not an identity is to plug in numbers that break the identity. That is, make the equation
3:13:00not true. Now, if you're just trying to decide if an equation is an identity or not, and
3:13:06not worried about proving it, then I recommend plugging in numbers, or graphing the left
3:13:12and right sides to see if those graphs are the same. Recall that an identity is an equation
3:13:19that holds for all values of the variable. This video states and proves three identities
3:13:25called the Pythagorean identities. The first one is the familiar cosine squared theta plus
3:13:33sine squared theta equals one. The second one says tan squared theta plus one equals
3:13:41secant squared theta. And the third one goes cotangent squared theta plus one equals cosecant
3:13:50squared theta. Let's start by proving that cosine squared theta plus sine squared theta
3:13:56equals one. I'll do this by drawing the unit circle with a right triangle inside it by
3:14:04the definition of sine and cosine, the x&y coordinates of this top point r cosine theta
3:14:12and sine theta, the high partners of my triangle is one, since that's the radius of my unit
3:14:18circle. Now the length of the base of my triangle is the same thing as the x coordinate of this
3:14:28point. So that's equal to cosine theta. The height of this triangle is the same thing
3:14:37as the y coordinate of this point. So that's sine theta. Now the Pythagorean theorem for
3:14:45right triangles says this side length squared plus that sine squared is equal to the hypothenar
3:14:52squared. So by the Pythagorean theorem, we have that cosine theta square plus sine theta
3:15:02squared equals one squared, I can rewrite that as cosine squared theta plus sine squared
3:15:09theta equals one, since one squared is one, and cosine squared theta is just a shorthand
3:15:15notation for cosine theta squared. That completes the proof of the first Pythagorean identity,
3:15:23at least in the case when the angle theta is in the first quadrant. In the case, when
3:15:28the angle was in a different quadrant, you can use symmetry to argue the same identity
3:15:34holds. But I won't give the details here. To prove the next Pythagorean identity, tan
3:15:40squared theta plus one equals secant squared theta, let's use the first without your an
3:15:45identity, which said that cosine squared theta plus sine squared theta equals one, I'm going
3:15:53to divide both sides of this equation by cosine squared theta. Now I'm going to rewrite the
3:16:01left side by breaking apart the fraction into cosine squared theta over cosine squared theta
3:16:08plus sine squared theta over cosine squared theta. Now cosine squared theta over cosine
3:16:15squared theta is just one. And I can rewrite the next fraction as sine of theta over cosine
3:16:23of theta squared. That's because when I square a fraction, I can just square the numerator
3:16:31and square the denominator. And sine squared theta is shorthand for a sine of theta squared.
3:16:36Similarly, for cosine squared theta. Now on the other side of the equal sign, I can rewrite
3:16:43this fraction as one over cosine theta squared. Again, that's because when I square the fraction,
3:16:50I just get the one squared, which is one divided by the cosine theta squared, which is this.
3:16:59I'm almost done. sine theta over cosine theta is the same thing as tangent theta. And one
3:17:06over cosine theta is the same thing as secant theta. Using the shorthand notation, that
3:17:14says one plus tan squared theta equals sequencer data, which, after rearranging is exactly
3:17:25the identity that we were looking for. The proof of the third for that green identity
3:17:31is very similar. Once again, I'll start with the identity cosine squared theta plus sine
3:17:37squared theta equals one. And this time, I'll divide both sides by sine squared theta. I'll
3:17:44break up the fraction on the left. And now I'll rewrite my fractions as cosine theta
3:17:51over sine theta squared plus one equals one over sine theta squared. Cosine over sign
3:18:01can be written as cotangent. And one over sine can be written as cosecant. That gives
3:18:13me the identity that I'm looking for. We've now proved three trig identities. The first
3:18:20one, we proved using the unit circle and the Pythagorean Theorem. The second and third
3:18:31identities, we proved by using the first identity and a bit of algebra. The sum and difference
3:18:38formulas are formulas for computing the sine of a sum of two angles, the cosine of a sum
3:18:46of two angles, the sine of a difference of two angles, and the cosine of a difference
3:18:52of two angles. Please pause the video for a moment to think about this question. Is
3:18:58it true that the sine of A plus B is equal to the sine of A plus the sine of B? No, it's
3:19:08not true. And we can see by an example, if we plug in say, A equals pi over two and B
3:19:16equals pi, then the sine of pi over two plus pi is the same thing as a sine of three pi
3:19:24over two, which is negative one. Whereas the sine of pi over two plus the sine of pi is
3:19:32equal to one plus zero, which is one, negative one is not equal to one. So this equation
3:19:40does not hold for all values of a and b. There are a few values of a and b for which it does
3:19:46hold. For example, if a is zero and B zero, but it's not true in general. Instead, we
3:19:54need more complicated formulas. It turns out that the sine of the sum of two ends angles
3:20:00A plus B is given by sine of A cosine of B plus cosine of A, sine of B. The cosine of
3:20:12A plus B is given by cosine A, cosine B minus sine A sine Bay. I like to remember these
3:20:23with a song, sine cosine cosine sine cosine cosine minus sine sine. Please feel free to
3:20:31back up the video and sing along with me, I encourage you to memorize the two formulas
3:20:37for the sine of a sum of angles and the cosine of a sum of angles. Once you do, it's easy
3:20:42to figure out the sine and cosine of a difference of two angles. One way to do this is to think
3:20:48of sine of A minus B as sine of A plus negative B. And then use the angle sum formula. So
3:20:59this works out two, sine cosine plus cosine, sine. And now, if I use the fact that cosine
3:21:12is even, I know that cosine of negative B is cosine of B. And since sine is odd, sine
3:21:20of negative b is negative sine of B. So I can rewrite this as sine of A cosine of B
3:21:31minus cosine of A sine of B. Notice that this new formula for the difference is the same
3:21:39as the formula for the sum is just that plus sign turned into a minus sign, we can do the
3:21:45same trick for cosine of A minus B, that's cosine of A plus minus b, which is cosine
3:21:54A cosine minus b minus sine of A sine of negative B. Again, using even an odd properties, this
3:22:05gives us cosine A cosine B plus sine A sine B. Once again, the formula for the difference
3:22:15is almost exactly like the for the sum, just that minus sign has switched to a plus sign.
3:22:21Now let's use the angle sum formula to find the exact value for the sine of 105 degrees.
3:22:28Now, 105 degrees is not a special angle on the unit circle, but I can write it as the
3:22:34sum of two special angles, I can write it as 60 degrees plus 45 degrees. Therefore,
3:22:42the sine of 105 degrees is the sine of 60 plus 45. And now by the angle sum formula,
3:22:52this is sine, cosine, cosine, sine. And I, from my unit circle, I can figure out that
3:23:04sine of 60 degrees is root three over two cosine of 45 degrees root two over two, cosine
3:23:11of 60 degrees is one half, and sine of 45 degrees is root two over two. So this simplifies
3:23:18to root six plus root two over four. For our last example, let's find the cosine of v plus
3:23:29W, given the values of cosine v, and cosine W, and the fact that v and w are angles in
3:23:35the first quadrant. Remember, to compute the cosine of a sum, we can't just add together
3:23:42the two cosines. That wouldn't even make sense in this case, because adding point nine and
3:23:46point seven would give something bigger than one and the cosine of something's never bigger
3:23:50than one. Instead, we have to use the angle sum formula for cosine. So that goes cosine
3:23:57of v plus w equals cosine, cosine, minus sine, sine. Now, I've already know the cosine of
3:24:09v and the cosine of W. So I could just plug those in. But I have to figure out the sine
3:24:13of v and the sine of W from the given information. And one way to do that is to draw right triangles.
3:24:20So here, I'm going to draw a right triangle with angle V, and another right triangle with
3:24:28angle W. Since I know that the cosine of V is point nine, I can think of that as nine
3:24:38over 10. And I can think of that as adjacent over hypotenuse in my right triangle. So I'll
3:24:45decorate my triangles adjacent side with the number nine and the hypotony is with 10. Similarly,
3:24:51since I know that the cosine of W is point seven, which is seven tenths, I can put a
3:24:56seven on this adjacent side and attend on this side. partners'. Now the Pythagorean
3:25:03Theorem, lets me compute the length of the unlabeled side. So this one is going to be
3:25:09the square root of 10 squared minus nine squared, that's going to be the square root of 19.
3:25:16And here I have the square root of 10 squared minus seven squared. So that's the square
3:25:20root of 51. I can now find the sine of V as the opposite over the hypotenuse. So that's
3:25:29the square root of 19 over 10. And the sine of W will be the square root of 51 over 10.
3:25:39Because we're assuming v and w are in the first quadrant, we know the values of sign
3:25:43need to be positive, so we don't need to Jimmy around with positive or negative signs in
3:25:48our answers, we can just leave them as is. Now we're ready to plug into our formula.
3:25:56So we have that cosine of v plus w is equal to point nine times point seven minus the
3:26:04square root of 19 over 10, times the square root of 51 over 10. using a calculator, this
3:26:12works out to a decimal approximation of 0.3187. This video gave the angle sum and difference
3:26:22formulas and use them to compute some values. To see a proof for why the sum formulas hold,
3:26:30please watch my other video. Remember the angle some formulas, those are the formulas
3:26:36for computing sine of A plus B, and cosine of A plus B. I like to sing them, sine cosine,
3:26:48cosine, sine, cosine, cosine minus sine sine. This video gives a geometric proof of those
3:27:02formulas. There are many great proofs of the angle some formulas, but I'd like to share
3:27:08with you one of my favorites for those who are interested. Alright, the angle some formulas
3:27:13up here, so we'll know what we're trying to prove. To prove these formulas, let me start
3:27:18by drawing an angle A and angle B on top of that. Next, I'm going to draw a line perpendicular
3:27:32to this middle line. And I'm going to extend the top line until it meets that perpendicular,
3:27:37making a right triangle. Finally, I'll draw a rectangle around that right triangle that
3:27:45just touches its vertices. My rectangle is now divided up into four right triangles.
3:27:52And I'm going to choose units of measurement, so that the high partners of my middle triangle
3:27:57has length one. Now let's stop for a minute to think about the angles of these other triangles.
3:28:06Since the top and the bottom edge of the rectangle are parallel lines, and this hypothenar is
3:28:13is a transversal. This angle up here must have the same measure as a plus b down here.
3:28:22Also, this skinny angle here must have the same measure as a down here because this angle
3:28:32is 180 degrees, minus 90 degrees minus this angle here. And this angle A is also 180 degrees,
3:28:41the measure of the angles in a triangle minus 90 degrees minus that same angle. So I'll
3:28:48label this skinny angle with a. Next, let's figure out as many side length as we can.
3:28:56Based on the middle right triangle with high partners one, we know that this side length
3:29:02down here must be cosine of B, since adjacent over hypotenuse is cosine B. Similarly, this
3:29:12side length here must be sine of B. Since opposite over hypotenuse is sine of B. Now
3:29:22we see that sine of B is the hypotenuse of this right triangle, which means that this
3:29:30little side here has measure sign a time sign B. That's because the opposite over the hypotenuse
3:29:43of this angle has to equal sign a. A similar argument shows that this side has to have
3:29:50measure cosine A time sign B. Please pause the video and take a moment to fill in the
3:29:57side length of this right triangle. In this right triangle, you should be getting sine
3:30:06A cosine B cosine A cosine B, sine of A plus B and cosine of A plus B. But remember, we
3:30:26have a rectangle here. So the opposite sides have equal length. This tells us that sine
3:30:35of A plus B has to equal sine of A cosine of B plus cosine of A sine of B, which is
3:30:42exactly the first angle sum formula. Also, cosine of A plus B, which is this side length,
3:30:51is exactly the difference of this side length cosine A cosine B minus this side length,
3:30:58sine of A sine B. And that's the second angle sum formula. So I think that's a pretty great
3:31:05geometric proof of the angle some formulas. This video gives formulas for a sine of two
3:31:13theta and cosine of two theta. Please pause the video for a moment and see if you think
3:31:20this equation sine of two theta equals two sine theta is true or false. Remember that
3:31:25true means always true for all values of theta, were false means sometimes they're always
3:31:32false. This equation is false, because it's not true for all values of theta. One way
3:31:39to see this is graphically, if I graph y equals sine of two theta, that's like the graph of
3:31:48sine theta, squished in horizontally by a factor of one half. On the other hand, if
3:31:55I graph y equals two sine theta, that's like the graph of sine theta stretched vertically
3:32:02by a factor of two. These two graphs are not the same. So instead, we need a more complicated
3:32:10formula for sine of two theta. And that formula is sine of two theta is two sine theta, cosine
3:32:21theta. It's not hard to see why that formula works based on the angle song formula. Recall
3:32:28that sine of A plus B is equal to sine A, cosine B plus cosine A, sine B. Therefore,
3:32:40sine of two theta, which is sine of theta plus theta is going to be sine theta, cosine
3:32:47theta, plus cosine theta sine theta. simply plugging in theta for a and theta for B, in
3:32:56this angle, some formula. Well, sine theta cosine theta is the same thing as cosine theta
3:33:02sine theta. So I can rewrite this as twice sine theta cosine theta. That gives me this
3:33:15formula. There's also a formula for cosine of two theta. And that formula is cosine squared
3:33:21theta minus sine squared theta. Again, we can use the angle sum formula to see where
3:33:30this comes from. cosine of A plus B is equal to cosine of A cosine A B minus sine A, sine
3:33:39B. So if we want cosine of two theta, that's just cosine of theta plus theta, which is
3:33:46cosine theta cosine theta, minus sine theta sine theta by plugging in beta for A and B.
3:33:58This can be rewritten as cosine squared theta minus sine squared theta, which is exactly
3:34:03the formula above. Now there are a couple other formulas for cosine of two theta that
3:34:09are also popular. One of them is one minus two sine squared theta. And the other one
3:34:17is cosine of two theta is two cosine squared theta minus one. You can get each of these
3:34:25two formulas from the original one using the Pythagorean identity. We know that cosine
3:34:33squared theta plus sine squared theta is one. So cosine squared theta is one minus sine
3:34:40squared theta. If I plug that into my original formula, which I've copied here, so I'm plugging
3:34:48in instead of cosine squared, I'm going to write one minus sine squared theta, I still
3:34:55have a nother minus sine squared theta. So that's the same thing as one minus Twice science
3:35:01grade data, which is exactly what I'm looking for. Similarly, I can use the Pythagorean
3:35:08identity to write sine squared theta as one minus cosine squared theta. Again, I'll take
3:35:16this equation and copy it below. But this time, I'm going to plug in for sine squared
3:35:24right here. So that gives me cosine of two theta is cosine squared theta minus the quantity
3:35:32one minus cosine squared theta. That simplifies to two cosine squared theta minus one after
3:35:40distributing the negative sign, and combining like terms. So I have one double angle formula
3:35:47for sine of two theta. And I have three versions of the double angle formula for cosine of
3:35:52two theta. Now let's use these formulas in some examples. Let's find the cosine of two
3:35:58theta. If we know that cosine theta is negative one over root 10, and theta terminates in
3:36:04quadrant three, we have a choice of three formulas for cosine of two theta, I'm going
3:36:12to choose the second one, because it only involves cosine of theta on the right side.
3:36:16And I already know my value for cosine theta. Of course, I could use one of the other ones,
3:36:22but then I'd have to work out the value of sine theta. So plugging in, I get cosine of
3:36:29two theta is twice negative one over root n squared minus one, which simplifies to two
3:36:37tenths minus one or negative eight tenths, negative four fifths. Finally, let's solve
3:36:45the equation two cosine x plus sine of 2x equals zero. What makes this equation tricky
3:36:52is that one of the trig functions has the argument of just x, but the other trig function
3:36:57has the argument of 2x. So I want to use my double angle formula to rewrite sine of 2x.
3:37:04I'll copy down the two cosine x, and now sine of 2x is equal to two sine x cosine x. At
3:37:14this point, I see a way to factor my equation, I can factor out a two cosine x from both
3:37:22of these two terms. That gives me one plus sine x and the product there is equal to zero.
3:37:34That means that either two cosine x is equal to zero, or one plus sine x is equal to zero.
3:37:42That simplifies two cosine x equals zero, or sine x is negative one. Using my unit circle,
3:37:50I see that cosine of x is zero at pi over two, and three pi over two, while sine of
3:38:00x is negative one at three pi over two, there's some redundancy here, but my solution set
3:38:07is going to be pi over two plus multiples of two pi, and three pi over two plus multiples
3:38:16of two pi. This video proved the double angle formulas, sine of two theta is two sine theta
3:38:23cosine theta. and cosine of two theta is cosine squared theta minus sine squared theta. It
3:38:32also proved two alternate versions of the equation for cosine of two theta. This video
3:38:40is about the half angle formulas for computing cosine of theta over two and sine of theta
3:38:47over two. Cosine of theta over two is either a plus or minus the square root of one plus
3:38:57cosine theta over two. To figure out whether to use plus or minus, you need to know something
3:39:04about what quadrant, the angle theta over two is, if you're expecting cosine theta over
3:39:09two to be negative, then you're gonna need the negative sign, you're expecting this cosine
3:39:14to be positive, then of course, you want the positive sign since the square root of something
3:39:18is always positive. Notice that this formula always makes sense, because the inside of
3:39:24the square root sign is always going to be positive or zero. Since cosine theta can't
3:39:29be any more negative than negative one, we have a similar formula for sine of theta over
3:39:35two, and that's plus or minus the square root of one minus cosine theta over two. Notice
3:39:43that the formulas for both cosine of A half angle and sine of A half angle both have cosine
3:39:50in them, they just differ by the positive or negative side and the inside of the square
3:39:55root. To see why these formulas hold, let's recall the form In the last four double angles,
3:40:02we know that cosine of 2x is equal to two cosine squared x minus one, or we can write
3:40:10cosine of 2x as one minus two, sine squared of x. Of course, of course, we can also write
3:40:18cosine of 2x as cosine squared x minus sine squared x. But this one won't be relevant
3:40:26to our discussion right now. So I'm going to start with this first formula. And I'm
3:40:32going to make the substitution theta equals to x, that allows me to write cosine theta
3:40:40is two cosine squared theta over two minus one. Since if theta is 2x, then x is theta
3:40:48over two. Now, I'm going to solve for cosine of theta over two. So I'll add one to both
3:40:58sides. Divide both sides by two and take the square root of both sides. That
3:41:11gives me that cosine of theta over two is going to be plus or minus the square root
3:41:17of cosine theta plus one over two, which is the formula we're looking for. I can carry
3:41:22a similar process out with the second formula. I'll substitute theta for 2x, which means
3:41:29that x is theta over two, that gives me cosine theta is one minus two sine squared theta
3:41:36over two. Now I'll solve for sine of theta over two. I'll start by adding two sine squared
3:41:46theta over two to both sides. Next, I'll subtract cosine theta from both sides. Next, I'll divide
3:41:56both sides by two. And finally, I'll take the square root of both sides. That gives
3:42:02me sine of theta over two is plus or minus the square root of one minus cosine theta
3:42:09over two, which is the formula I was looking for. Now that we've got our formulas, and
3:42:15know where they come from, let's use them in an example. Let's suppose that sine theta
3:42:21is four fifths, and theta is between pi over two and pi. That means it's in the second
3:42:27quadrant, we want to find the exact values of cosine theta over two and sine theta over
3:42:34two. Since theta is between pi over two and pi, theta over two will be between pi over
3:42:41four and pi over two. So theta over two is somewhere in the first quadrant. That means
3:42:51that cosine theta over two and sine theta over two will both be positive. Let me write
3:42:57down my half angle formulas. And I know I can ignore the negative version and just use
3:43:03the positive version. Now unfortunately, I'm just given the value of sine theta and not
3:43:08cosine theta, so I can't plug in directly, first, I have to use this value to figure
3:43:12out what cosine might be. I'll draw a right triangle with angle theta. And since sine
3:43:21of theta is four fifths, I'll decorate the triangle with a four and a five on the hypothesis.
3:43:28That means that this side length is going to be five squared square root of five squared
3:43:32minus four squared, which is the square root of nine, which is three. That means that cosine
3:43:40theta is going to be either plus or minus three fifths because it's adjacent over hypotenuse.
3:43:50Now since I'm in the angle, the second quadrant for my angle theta, cosine is actually going
3:43:55to be negative three fifths. Now I can plug into my formulas, I get that cosine of theta
3:44:03over two is going to be the square root of one minus three fifths over two. That simplifies
3:44:10to the square root of two fifths divided by two, which is the square root of 1/5, or one
3:44:15over the square root of five. Sine of theta over two is the square root of one minus cosine
3:44:24theta, so that's one minus negative three fifths over two, that's the square root of
3:44:33one plus three fifths, or eight fifths over to the square root of four fifths or two over
3:44:39the square root of five. In this video, we found formulas for the cosine of A half angle,
3:44:46and the sine of A half angle and use them in an example. solving a right triangle means
3:44:53finding the lengths of all the sides and the measures of the angles given partial information.
3:45:02In this example, we're given the length of one side, and the measure of one angle. Plus
3:45:08we know the measure of this right angle is 90 degrees, we need to find the measure of
3:45:14the third angle labeled capital A, and the length of the two sides labeled lowercase
3:45:20b and lowercase C. To find the measure of angle A, let's use the fact that the measures
3:45:27of the three angles of a triangle add up to 180 degrees. So that means that 49 degrees
3:45:34plus 90 degrees plus a is equal to 180 degrees. So A is equal to 180 degrees minus 90 degrees
3:45:45minus 49 degrees, which works out to 41 degrees. To find the length of the side B, we have
3:45:53a couple of possible options. We could use the fact that tan of 49 degrees, which is
3:46:01opposite over adjacent is B over 23. So b is 23 times tan 49 degrees, which works out
3:46:12to 26.46 units. Alternatively, we could use the fact that tan of 41 degrees is 23 over
3:46:23b since now if we're looking at the angle here, 23 is our opposite and B is an adjacent,
3:46:31that's a little bit harder to solve algebraically. But we can write B tan 41 degrees equals 23,
3:46:38which means that B is 23 divided by 1041 degrees with a calculator that works out again to
3:46:4726.46. The reason we want to use tan in this problem and not say sine or cosine is because
3:46:54tan of say 49 degrees relates the unknown side that we're looking for B to the side
3:47:01that we know the measure of if we had use sine instead, we'll be saying that sine of
3:47:0749 is B over C and we'd have two unknowns, which would make it difficult to solve. Next,
3:47:15to find the side length C, we can have a few options, we could use a trig function again.
3:47:21For example, we could use the cosine of 49 degrees, that's adjacent over hypotenuse,
3:47:31which is 23 oversee. Solving for C, we get that C is 23 over cosine 49. Which works out
3:47:43to 35.06 units. Another option would be to use the Pythagorean Theorem to find C. Since
3:47:53we know 23 squared plus b squared equals c squared. In other words, that's 23 squared
3:48:00plus 26.46 squared equals c squared, which means that C is the square root of that song,
3:48:11which works out again to 35.06. To review the ideas we used where the sum of the angles
3:48:22is equal to 180 degrees. We used facts like tangent of an angle being opposite over adjacent
3:48:30and similar facts about sine and cosine. And we use the Pythagorean Theorem. This allowed
3:48:36us to find all the angles and side lengths of the triangle knowing just the side length
3:48:41of one side and the angle of one of the non right angles to begin with. In this next example,
3:48:51we don't know any of the angles except for the right angle, but we know to have the side
3:48:56lengths. To find the unknown angle theta, we can use the fact that cosine theta is adjacent
3:49:03overhype hotness, so that's 10 over 15. Cosine is a good trig function to use here, because
3:49:11this equation relates our unknown angle to our two known sides. So we just have one unknown
3:49:18in our equation to solve for. To solve for theta, we just take the cosine inverse of
3:49:2410/15 which is 0.8411 radians, or 48.19 degrees. To find the measure of angle fee, we could
3:49:40use the fact that sine of fee is 10 over 15 and take sine inverse of 10/15. But probably
3:49:46a little easier just to use the fact that these three angles do 180 degrees. That tells
3:49:52us that fee plus 90 plus 48.19 is equal to 180 Which means that fee is 41.81. Finally,
3:50:09we can find x, either using a trig function or by using the Pythagorean Theorem. To find
3:50:16it using a trig function, we could write down something like tan of 48.19 degrees is x over
3:50:2410. To find that using the Tyrion theorem, we'd write down 10 squared plus x squared
3:50:29equals 15 squared. I'll use the Pythagorean theorem and find the x by doing the square
3:50:36root of 15 squared minus 10 squared. That gives me an answer of 11.18. Notice that we
3:50:45use many of the same ideas as in the previous problem. For example, the fact that the sum
3:50:50of the angles is 180. The Pythagorean theorem and the trig functions like tan, sine and
3:50:59cosine, we also use the inverse trig functions to get from an equation like this one to the
3:51:06angle. This video showed how it's possible to find the lengths of all the sides of a
3:51:11right triangle. And then measures of all the angles given partial information. For example,
3:51:17the measure of one angle and one side or from two sides. Recall that solving a triangle
3:51:27means finding all the lengths of the sides and all the measures of the angles from partial
3:51:33information. The Law of Cosines is a tool that can help us solve triangles that are
3:51:39not necessarily right triangles. Recall that the Pythagorean theorem says that for a right
3:51:46triangle, like this one, the length of the hypothesis is related to the lengths of the
3:51:53sides by the formula c squared equals a squared plus b squared. I like to think of the law
3:52:00of cosines. As a generalization of the Pythagorean Theorem, two triangles that are not necessarily
3:52:06right triangles. loosely speaking, the Law of Cosines says that c squared is equal to
3:52:13a squared plus b squared plus a correction factor where the correction factor depends
3:52:20on the size of the angle, that's opposite to side C. For this first triangle, on the
3:52:29left, the angle that's opposite to side C is exactly a right angle. So we know that
3:52:36c squared is exactly equal to a squared plus b squared. But in this next triangle, the
3:52:45angle opposite decides C prime is a little bit less than a right angle. So we expect
3:52:52the side length C prime should be a little bit shorter than the side length C in the
3:52:57previous triangle that had the same side length A and B for its legs. Therefore, C prime squared
3:53:05should be less than a squared plus b squared, or in other words, C prime squared should
3:53:11equal a squared plus b squared minus a little bit. In this third triangle, the angle opposite
3:53:19C double prime is a little bigger than 90 degrees. So this side length C double prime
3:53:25should be a little bit longer than the side length C in the right triangle with the same
3:53:30size legs. So C double prime squared should be bigger than a squared plus b squared, we
3:53:39can write this as C double prime squared is equal to a squared plus b squared plus a little
3:53:45bit. The Law of Cosines says precisely what this little correction factor is. It says
3:53:52that for any triangle with sides, a B and C, and angle capital C, opposite side C, side
3:54:00c squared is equal to a squared plus b squared minus two A B, cosine angle C. Let's analyze
3:54:11what happens with this equation when angle C is equal to 90 degrees less than 90 degrees
3:54:17and greater than 90 degrees. If angle C is equal to 90 degrees, then cosine of angle
3:54:26C is equal to zero. So we have that c squared is equal to a squared plus b squared, the
3:54:32ordinary Pythagorean Theorem. If on the other hand, angle C is less than 90 degrees, then
3:54:39from the unit circle, we know that cosine of C is positive. So to a B, cosine C is a
3:54:48positive number, and we'll be subtracting a little bit from our A squared plus B squared,
3:54:54just like we saw in the picture. Finally, if angle C is bigger than 90 degrees, then
3:55:01we can see from the unit circle that cosine FC is negative. So to a B, cosine capital
3:55:09C is going to be less than zero. And we're going to be subtracting a negative number,
3:55:15which means we're actually adding a little bit, just like we saw in the third picture.
3:55:21When labeling a triangle, the convention is to use capital letters for the angles and
3:55:27lowercase letters for the side lengths. And to put side A opposite angle A, and so on.
3:55:34When we wrote the law of cosines, on the previous page, we wrote c squared equals a squared
3:55:40plus b squared minus two A B, cosine angle C. But it doesn't matter which side we call
3:55:48a, which side we call B, and which side we call C. All that matters is that this angle
3:55:53is opposite to the side and between these two sides. So we could have just as easily
3:56:00written a squared is equal to b squared plus c squared minus two B, C, cosine A. all we've
3:56:09done is replace C with A, A with B, and B with C and our labeling scheme. Or instead,
3:56:21we could put the B squared on the left side and right b squared equals a squared plus
3:56:26c squared minus two a C, cosine B, we can use whichever of these three forms is most
3:56:33convenient for the problem at hand. Let's use the law of cosines. To find all the side
3:56:39lengths and angles in this triangle. By convention, I'll call the side opposite to angle A little
3:56:46a and the side opposite angle B, little B. Notice that we know two side lengths and the
3:56:58angle between them. So that's an S, a s, side angle side triangle. If we write down the
3:57:07law of cosines, with the unknown side, c squared on the left side, then we have all the values
3:57:15for the variables on the right side. So let's plug them in. I'll use a calculator in degree
3:57:21mode to find that c squared is equal to 44.44. taking the square root, I get that c is equal
3:57:30to 6.67. Next, I'll use the law of cosines. To find the angle B, I'll need to use a version
3:57:39of the Law of Cosines that has cosine be right here. That's the version that has little b
3:57:46squared on the left side and goes b squared equals a squared plus c squared minus two
3:57:54a C, cosine B. I can plug in values for all my side lengths. And now I can solve for cosine
3:58:04B. To do that, I'll subtract eight squared and 6.67 squared from both sides. And now
3:58:12I'll divide both sides by negative two times eight times 6.67. using a calculator, this
3:58:19gives me that cosine of B is equal to negative 0.1172. The negative value of cosine indicates
3:58:29that my angle B must be greater than 90 degrees, which agrees with the figure. Next, I can
3:58:37take cosine inverse of both sides to get that b is equal to cosine inverse negative 0.1172,
3:58:46which works out to 96.73 degrees. The last thing I need to do is solve for angle A, I
3:58:55could use the Law of Cosines again, and work out angle A just like I worked out angle B
3:59:00here. But a simpler thing to do is to use the fact that the sum of the three angles
3:59:06equals 180 degrees. In other words, a plus 37 degrees plus 96.73 degrees equals 180 degrees.
3:59:17So A is going to be 180 minus 37 minus 96.73, which works out to 46.27 degrees. In the previous
3:59:31example, we were given two sides, and the angle between them. In this example, instead,
3:59:37we're given three side lengths. So I'll call this and s, s, s, side side side triangle,
3:59:46we need to find all three angles. Although there are a lot of computations involved,
3:59:51the ideas are the same as in the previous problem. Defined angle capital C, we need
3:59:58the form of the Law of Cosines That has cosine of C on the right side, and little C on the
4:00:05left side. To find angle A, we need the form of the Law of Cosines that has cosine A on
4:00:13the right side and little a on the left side. And to find angle B, we need the form of the
4:00:19Law of Cosines that has cosine of B on the right side, and little b squared on the left
4:00:27side. For each of these three equations, I'll plug in the side lengths. So for the cosines
4:00:36and use inverse cosine to find my angles. Notice that my three angles 60 degrees, 32
4:00:44point 20 degrees, and 87.79 degrees add up to just almost 180 degrees, the difference
4:00:53is just around off error. In fact, I could have saved myself some work by just finding
4:00:59the measure of angle C, and the measure of angle A, and then subtracting them from 180
4:01:04degrees to get the measure of angle B. In this video, we stated the law of cosines and
4:01:12used it to solve some triangles. The Law of Cosines can be thought of as the Pythagorean
4:01:18Theorem, with the correction factor to account for triangles that aren't right triangles.
4:01:24Recall that solving a right triangle means finding all the lengths of the sides and all
4:01:30the measures of the angles from partial information. The Law of Cosines is a tool that can help
4:01:36us solve triangles that are not necessarily right triangles. We call it the Pythagorean
4:01:42theorem says that for a right triangle, like this one, c squared is equal to a squared
4:01:51plus b squared. I like to think of the law of cosines. As a generalization of the Pythagorean
4:01:57Theorem, two triangles that are not necessarily right triangles. loosely speaking, the Law
4:02:03of Cosines says that c squared is equal to a squared plus b squared plus a correction
4:02:11factor where the correction factor depends on the size of the angle that's opposite sides
4:02:18see, in this first triangle, on the left, the angle opposite to side C is a right angle.
4:02:25So we know that c squared is equal to exactly a squared plus b squared. But on the next
4:02:34triangle, the angle opposite to side C prime is a little bit less than a right angle. So
4:02:42this side, C prime should be a little shorter than a squared plus b squared. All right,
4:02:49C prime squared is equal to a squared plus b squared minus a bit. In this third triangle,
4:03:00says the angle opposite to side C double prime is a little bit bigger than 90 degrees, this
4:03:07side length should be a little bit bigger than a squared plus b squared. So I'll write
4:03:12it as a squared plus b squared plus a little bit. The Law of Cosines says exactly what
4:03:19this little correction factor is. It says that for any triangle with sides, A, B and
4:03:24C and angle capital C are opposite to side C. C squared is equal to a squared plus b
4:03:31squared minus two A B, cosine of angle C. No notice that if the angle C is less than
4:03:4190 degrees, cosine of C is going to be a positive number. And so we'll be subtracting a positive
4:03:51quantity, just like we saw in the picture above. If however, angle C is bigger than
4:03:5890 degrees, then cosine of c is negative. So by subtracting to a B, cosine C, we're
4:04:06actually adding a little bit, and we get a longer side as in this picture. When labeling
4:04:13a triangle, the convention is to use lowercase letters for the side lengths and uppercase
4:04:19letters for the angles, and to put angle A opposite side A, and so on. When we wrote
4:04:26the law of cosines, on the previous page, we wrote c squared equals a squared plus b
4:04:31squared minus two A B, cosine C. But it doesn't matter which side we call a, which side we
4:04:40call B, and which side the coffee. All that matters is that this angle is opposite to
4:04:46this side, and between these two sides, so we could have just as easily written a squared
4:04:54equals b squared plus c squared minus two B, C code sine A or B squared equals a squared
4:05:05plus c squared minus two a C, cosine B, we can use whichever of these three forms is
4:05:13most convenient for the problem at hand. Let's use the law of cosines. To find the side length
4:05:20the angles of this triangle, by convention, I'm going to call the side opposite angle
4:05:26B, side, little B, and the side opposite angle A side, little a. Since we know two side lengths,
4:05:38and the angle between them, we have all the information on the right side of the law of
4:05:43cosines. So I'll plug that in. And use a calculator in degree mode to find that c squared equals
4:05:5244.44. taking the square root, I get that C equals 6.67. Next, I'll use the law of cosines.
4:06:05To find the angle B, I need to use a version of the law of cosines. That mentions cosine
4:06:12of angle B here, so it'll need to mention side length little B here. So that'll be the
4:06:18form b squared. On the left, I'll get a squared plus c squared minus two a C, cosine B. On
4:06:29the right, I'll plug in values from my side lengths. And now I can solve for cosine B
4:06:36by subtracting eight squared, and 6.67 squared from both sides. And now dividing both sides
4:06:44by negative two times eight times 6.67. using a calculator, this gives me that cosine of
4:06:51B is equal to negative point 1172. The negative value of cosine indicates that my angle B
4:07:02must be greater than 90 degrees, which agrees with a picture. Next, I can take cosine inverse
4:07:08of both sides to get that b is equal to cosine inverse of negative point 1172, which works
4:07:19out to 96.73 degrees. The last thing I have to do is solve for angle A, I could use the
4:07:29Law of Cosines again, and work it out just like I did here for angle B. But a simpler
4:07:35method is to just use the fact that the sum of the three angles is 180 degrees. So a must
4:07:43be 180 degrees minus 37 degrees minus 96.73 degrees. This works out to 46.27 degrees.
4:07:58In the previous example, we were given two sides and the angle between them. In this
4:08:03angle, were given instead, three side lengths, and we need to find all three angles. Although
4:08:09there are a lot of computations involved, the ideas are the same as in the previous
4:08:14problem. To find angle capital C, we need the form of the law of cosines, that has cosine
4:08:21of capital C on the right side, to find capital B, we need to use the form of the Law of Cosines
4:08:27that has cosine of B on the right side, and little B on the left side. So it'll be b squared
4:08:33equals a squared plus c squared minus two a C cosine B. At the find angle A, we need
4:08:41a little A squared on the left side. So that will get cosine of angle A on the right side.
4:08:50For each of these three equations, I'll plug in the side lengths, solve for the cosine
4:08:56of the angle and use inverse cosine to find my angles. Notice that my three angles 60
4:09:04degrees, 87.79 degrees, and 32 point 20 degrees add up to almost exactly 180 degrees, it's
4:09:12just a tiny bit off due to roundoff error. In fact, I could have saved myself some work
4:09:18by just finding the measure of angle C, and the measure of angle B, for example, and then
4:09:23subtracting their sum from 180 degrees to get angle A. In this video, we stated the
4:09:30law of cosines and used it to solve some triangles. The Law of Cosines can be thought of as the
4:09:37Pythagorean Theorem with an adjustment factor to account for triangles that aren't right
4:09:42triangles. The Law of Sines is a tool that can be used to solve triangles that are not
4:09:48necessarily right triangles. It's especially useful when we know two angles and one side
4:09:54and it can also be used in this situation where we know two sides and an angle Let's
4:10:00not between them. The Law of Sines says that for a triangle with angles A, B and C, opposite
4:10:08two sides, lowercase a, b, and c respectively. The sine of angle A divided by the length
4:10:18of side A is equal to the sine of B over B, which is also equal to the sine of C over
4:10:26C. Notice that we're taking the sine of the three angles, and dividing by the three opposite
4:10:34side lengths. Suppose we know that angle A is 55 degrees, angle C is 67 degrees, and
4:10:43Side B is 20 degrees, we want to solve the triangle, that is find the length of the other
4:10:50sides and the measure of the third angle. Since we know two angles already, we can find
4:10:56the third angle by subtracting from 180 degrees. So angle B is 58 degrees. Now I can use the
4:11:05Law of Sines to find my remaining side lengths. To find side A, I'll use the fact that sine
4:11:15of angle A over side length A is equal to sine of angle B over side length B. Since
4:11:23I already know three of these quantities, angle A, angle B, and side length B, I can
4:11:31plug in and solve for the unknown side length a to solve or multiply both sides by a and
4:11:39then multiply both sides by 20 over the sine of 58 degrees. That gives me that a is equal
4:11:46to 20 sine 55 oversized 58, which works out to 19.32. Please pause the video and use the
4:11:59same method to find side length See, this time, we need to involve sign of C and sidelink
4:12:09see in our equation. And we could use either the B's or the A's since they're both known
4:12:16for the other side. I'll plug in values as I can and solve for C. C works out to 21.71.
4:12:29In the previous example, we use the Law of Sines to find the side lengths of a triangle,
4:12:35because we already knew the angles. In this next example, we're given some side lengths.
4:12:41And we're going to need to solve for some angles. This can be tricky, because when we
4:12:47write down the law of sines and start solving for an angle, we first solve for sine of the
4:12:55angle. But there can be two plausible angles that both have the same sign. For example,
4:13:02if sine of the angle happens to be one half, then the angle could be either 30 degrees,
4:13:12or 150 degrees. In some cases, this can lead to two possible triangles that both have the
4:13:21conditions given When this happens, and there are two possible solutions. I call this an
4:13:26ambiguous case. With that warning, let's figure out the details of this example. Plugging
4:13:33in my given information into the law of signs, I'll focus on this portion of the equation
4:13:40where there's only one unknown to solve for. First off, solve for sine of A, which is eight
4:13:47times sine 40 degrees over seven, which works out to 0.7346. From the unit circle, I can
4:13:59see that there'll be two possible angles with this sine one in the first quadrant and one
4:14:05in the second quadrant. If I take sine inverse of both sides of my equation, I get the solution
4:14:13in the first quadrant, which is 47.27 degrees. Since this angle right here is 47.27 degrees,
4:14:26so is this angle here. And so the angle that I want in the second quadrant I can get by
4:14:31going 180 minus 47.27 degrees, which works out to 132.731 of these possible angles. It's
4:14:44an acute angle, the other one and obtuse angle, but both are possible configurations are sketch
4:14:50the possible triangles in both triangles, side length A and B are the same and angle
4:14:59B is the same But everything else is different. We can continue to solve for the other angle
4:15:05and side in each case. In the left side, I'll call this case one, we can solve for angle
4:15:14C, by taking 180 degrees minus angle B minus angle A. That gives us an angle just over
4:15:2690 degrees 92.73 degrees. Now we can use either the Law of Sines or the law of cosines. To
4:15:33find the final side length C, I'll use the Law of Sines again. So I plug in the value
4:15:40of angle C into this part of my equation, and solve for C to get C equals sine 92.73
4:15:50degrees times seven over sine 40 degrees, which works out to 10.88. For the right hand
4:16:02triangle, we can carry out the same steps to get angle C. And now plug in to the Law
4:16:08of Sines. To get side length to see, we finish solving for the two possible triangles. With
4:16:18these given side length, and angle. In this video, we use the Law of Sines to solve a
4:16:25triangle in two different situations. In the first situation, we had two angles and the
4:16:30side between them. So that's an A s, a triangle, for angle, side angle. And that situation,
4:16:41we ended up just using the law of science to solve for side lengths, and there was no
4:16:47ambiguity. In the second situation, we had a triangle where we knew two side lengths
4:16:52and an angle that was not between them. So I'll call that s s A. And in that case, we
4:16:58did find two possible triangles. That would work. The SSA situation doesn't always give
4:17:06two possible triangles. Sometimes when you go through the two cases, and figure out all
4:17:12the other angles and side lengths, you'll get an impossible situation. And the angle,
4:17:17for example, that's got to be negative. Because the other two angles add up to more than 180
4:17:24degrees by but you do have to be on the lookout for two possible triangles when you're solving
4:17:30a triangle with the SSA information given. This video describes some of the features
4:17:36of a parabola. Its vertex, focus and directrix. I want to start by finding the equation of
4:17:44all points that are equidistant the same distance from a point zero p on the y axis, and the
4:17:52horizontal line y equals negative P. I'm assuming here that P is positive, for starters, certainly
4:18:01the origin will be among those points as it has a distance of P from the point and P from
4:18:08the line. But the other points on the x axis will be closer to the line than they are to
4:18:14the point. So if I want the set of points that are the same distance from the point
4:18:18of the line, that's going to be a curve that curved upwards, something like this. So for
4:18:25example, a point out here will be the same distance from the point as it is from the
4:18:31line. So our intuition is suggesting that this set of points should be the shape of
4:18:38a parabola. Let's confirm this with some algebra. If we take an arbitrary point, with coordinates
4:18:46x, y, its distance from the point zero P is given by the distance formula, x minus zero
4:18:54squared plus y minus p squared. Its distance from the line y equals negative P is just
4:19:03given by its difference in Y coordinates. So that's going to be y minus negative p or
4:19:14y plus P. Let me set these two quantities equal to each other. And simplify, I can square
4:19:22both sides to get on the left x minus zero squared with the same thing as x squared plus
4:19:29y minus p squared equals y plus p squared. Now I'll distribute out, that gives me x squared
4:19:39plus y squared minus two p y plus p squared equals y squared plus two p y plus p squared.
4:19:49The P squares cancel out as do the Y squared, and I'm left with x squared equals four p
4:19:58y after move In this negative two p y to the other side, I could also write this as y equals
4:20:06one over four p x squared. And you might recognize this as the standard parabola y equals x squared,
4:20:14transformed by vertically stretched it stretching or shrinking it by this factor of whenever
4:20:20for pee. You might recall that this lowest point of the parabola is called the vertex.
4:20:26This point here is called its focus. And the line here is called the directrix. Notice
4:20:36that the number of P in this equation represents the distance between the vertex and the focus,
4:20:43and also represents the distance between the vertex and the directrix line. For this reason,
4:20:49if you're interested in the focus and the vertex, this form of the equation might be
4:20:54more useful than the equivalent form something like y equals A x squared, where the that
4:21:01distance P is more hidden. Now, so far, we've just considered the case when a p is greater
4:21:07than zero. Let's look for a moment at what happens when p is less than zero, then when
4:21:12we draw the point zero, P is going to be below the x axis. And the directrix y equals negative
4:21:21P is going to be above the x axis because we're taking the negative of a negative number
4:21:25makes it a positive y value. Our parabola will also be upside down. All the algebra
4:21:35works out the same is just in our equation, this coefficient of y will now be a negative
4:21:41number instead of a positive one. Or if we write it in this other form, which might be
4:21:45more familiar, our coefficient of x squared will be negative instead of positive, which
4:21:50signifies that the parabola is pointing down. Since P is now a negative number, if we want
4:21:57to talk about distances, we should write absolute value of p instead of P. Next, let's turn
4:22:04things sideways by looking at the equation for all points equidistant from a point P
4:22:10zero. on the x axis. Here, we're assuming For starters, that P is bigger than zero,
4:22:16and a vertical line at x equals negative P. Our intuition suggests that this should be
4:22:24a parabola pointing sideways. And we can work out the same algebra steps to get an equation
4:22:30for it. Starting with an arbitrary point, x y, the distance from the point P zero is
4:22:40going to be given by x minus p squared plus y minus zero squared, the distance from the
4:22:46line x equals negative P is going to be given by the difference in x coordinates. So that's
4:22:54going to be x minus negative p, or just x plus P. I'll set these two distances equal
4:23:03to each other, square both sides like before, and simplify. This gives me the equation y
4:23:10squared equals four p x, which I could also rewrite as x equals one over four p y squared
4:23:18or x equals A y squared, where my a corresponds to my one over four P. Once again, if p is
4:23:26negative, then I probably pointing the opposite direction, but my equation will still be the
4:23:33same, it'll just have a negative coefficient of x instead of a positive one. The negative
4:23:38number of being hidden inside this variable P. As before, the absolute value of p gives
4:23:46me the distance between the vertex and the focus, and also gives me the distance between
4:23:53the vertex and the directrix. Now let's find the equation of a parabola with a vertex at
4:23:59some arbitrary point h k instead of the origin, assuming that the parabola opens up or down.
4:24:08If the parabola had its vertex at the origin, we know it would be given by the equation
4:24:13x squared equals four p y, or I could rewrite that as y equals one over four p x squared.
4:24:22So this parabola should just be shifted to the right by H and up by K. From transformations
4:24:29of functions. We know we can accomplish this by sticking a x minus h on the inside and
4:24:39putting a plus k on the outside. In other words, we have y minus k equals one over four
4:24:47P times x minus h squared. Or we can write this as x minus h squared equals four P times
4:24:57y minus k. So this is the original All with a vertex at the origin, and here is the transformed
4:25:08by shifting right and up to have a vertex at h k, notice that the equations are exactly
4:25:17the same. It's just you stuck in a minus h and a minus k near the X and near the y, this
4:25:27transform parabola will have its focus, a distance of P up from the vertex. So that's
4:25:32going to be the point h, k plus p, that's going to have as directrix at a distance p
4:25:40down from the vertex. So that's going to be the line y equals k minus P. Everything's
4:25:49the same for this other parabola pointing downwards, we just have P is negative in that
4:25:55case, the same story holds for the parabola is pointing right and left, the original equation,
4:26:03y squared equals four p x gets transformed when we shift the parabola over into y minus
4:26:11k squared equals four p x minus h. This time, the vertex is a distance p to the left or
4:26:20right of my vertex. So here it would be h plus p, k. And the directrix would be shifted
4:26:29over by P also from the vertex. So this would be the vertical line x equals h minus P. It
4:26:37works the same way, when the parabola is opening, left instead of right, it's just in this case,
4:26:43P is negative. Let's use this information to find the equation of a parabola with vertex
4:26:50at two, four, and focus at negative one, four. I'll start with a quick sketch, which will
4:26:56help me see how the parabola is lined up. So here's the vertex at two, four. And the
4:27:03focus is at negative one, four. So here's the focus here. Now says the problem always
4:27:11has its focus, that means that this problem must be pointing left and looks something
4:27:17like this. The directrix is going to be on the other side of the parabola, the same distance
4:27:23as the distance between the vertex and the focus. Since this distance is a distance of
4:27:29three to minus negative one is three, the directrix must be three units over here to
4:27:36the right, and so that would be the line x equals five. Since my problems open to the
4:27:47left, I know I need to use the form of the equation y minus k squared equals four p x
4:27:54minus h, where h k is the vertex, so that's two four, I also know that P is going to be
4:28:01negative. And since the absolute value of p represents the distance between the vertex
4:28:07and the focus, I know that P must be negative three. So filling on all these numbers, I
4:28:14get y minus four squared equals four times negative three times x minus two. I'll simplify
4:28:24that to y minus four squared equals negative 12x minus two. In this video, we looked at
4:28:31special forms of the equation for a parabola, the forms y minus k squared equals four p
4:28:38x minus h, and x minus h squared equals four p y minus k. In these equations, h k represents
4:28:47the vertex and PS absolute value gives the distance between the vertex and the focus
4:29:02and between the vertex and the directrix. parabolas in this form, are opening right
4:29:10if p is bigger than zero, and left if p is less than zero, and parabolas in this form,
4:29:17are opening up if p is bigger than zero, and down if p is less than zero. This video is
4:29:25about ellipses. Recall that a circle can be defined as a set of points whose distance
4:29:32from a fixed point is a constant. And he lips can be defined as a set of points, such that
4:29:41the sum of the distances from two fixed points is a constant. The two fixed points are called
4:29:55focuses or foe sigh The sum of the distances to the to fo Sai is the total length of the
4:30:05string, which is held constant as I draw the ellipse. Let's examine the features of an
4:30:10ellipse in more detail. In these pictures, the two black dots are the focuses or fo sigh.
4:30:19The red line segment that cuts through the middle of the ellipse in the long direction
4:30:24and passes through this low sigh, it's called the major axis. The line segment that cuts
4:30:30through the middle of the ellipse in the shorter direction is called the minor axis. And the
4:30:35two points at the tips of the ellipse where the major axis actually touches the ellipse
4:30:40are called the vertices. And ellipse could be elongated in any direction, but will only
4:30:47consider ellipses that are elongated either in the horizontal direction, or the vertical
4:30:52direction. Let's find the equation of an ellipse whose fo cy are at negative c zero, and c
4:31:01zero, and whose vertices are at negative a zero and a zero. This will be an ellipse that's
4:31:11elongated in the horizontal direction. For any point x, y on the ellipse, the sum of
4:31:19the distance from x y to the first focus, plus its distance to the second focus has
4:31:25to be some constant. And in fact, it turns out that constant has to equal to a, I'll
4:31:32show you why. If you look at this point right here on the far right tip of the ellipse,
4:31:39its distance from the first focus is going to be to C plus a minus C. And as distance
4:31:49from the second focus is just going to be a minus say, if you add up c plus c plus a
4:31:56minus c plus a minus C, you get exactly to a. So let me write out my formulas for distances,
4:32:05the distance from a point x y to the first focus, the point negative c zero is going
4:32:10to be the square root of x minus negative c squared plus y minus zero squared. And the
4:32:19distance from x y to the second focus, the point c zero is going to be the square root
4:32:25of x minus c squared plus y minus zero squared. That Psalm needs to add up to two a, after
4:32:32a fair amount of algebra, this simplifies the expression a squared minus c squared x
4:32:40squared plus a squared y squared equals a squared times a squared minus c squared. If
4:32:48we let b squared equals a squared minus c squared, then we can rewrite this as b squared
4:32:54x squared plus a squared y squared equals a squared b squared, and dividing both sides
4:33:01by a squared b squared, this gives us the form of the equation for the ellipse x squared
4:33:07over a squared plus y squared over b squared equals one. Since we got b squared as a squared
4:33:15minus c squared, it follows that b is less than a. In fact, B is half the length of the
4:33:22minor axis. In other words, this point right here is the point zero, B, and this point
4:33:29is the point zero negative B. To see why this is true, let's draw these two right triangles.
4:33:37The sum of the side lengths that I've drawn has to be the total length of the string.
4:33:43So that equals to a, which means each of these side lengths must be a. Now the base of this
4:33:50triangle is C. So by the Pythagorean Theorem, this height squared must be a squared minus
4:33:57c squared. Since b squared is defined as a squared minus c squared, this height must
4:34:03be B. To summarize, for a number a bigger than B, the equation x squared over a squared
4:34:10plus y squared over b squared equals one represents any lips. That's elongated in the horizontal
4:34:22direction, whose major axis terminates in the points minus a zero, and a zero, and whose
4:34:31minor access terminates in the points zero B, and zero minus b. It's fo cy are at the
4:34:41points negative c zero, and c zero. And the relationship between a B and C is that b squared
4:34:49is a squared minus c squared equivalently. If we add c squared to both sides, and subtract
4:34:57b squared from both sides, we can write See squared equals a squared minus b squared.
4:35:03The way I remember this is that the biggest of the three numbers A, B, and C is definitely
4:35:07going to be a, since a is half the length of the major axis, which is bigger than half
4:35:13the length of the minor axis, and definitely bigger than the length between the origin
4:35:18and one of the focuses. Since a is the biggest, and we have a Pythagorean Theorem relationship
4:35:24between a, b and c, the equation must be b squared plus c squared equals a squared, which
4:35:29can be rearranged to either of these two equations. Now what happens instead, if we have a bigger
4:35:36than B still, but this time, we're dividing y squared by the bigger number squared, and
4:35:41x squared by the smaller number squared? Well, this basically reverses the roles of x and
4:35:46y in my equation. And so we end up with the major axis on the y axis stretching between
4:35:52the points, zero a, and zero negative a, whereas the minor axis is going to be along the x
4:35:59axis going in between negative b zero, and B zero, and the first side are going to be
4:36:06here on the y axis again. So in this case, my lives is elongated in the vertical direction.
4:36:16Our previous ellipses were centered at the origin. What if the ellipse is centered instead
4:36:21of an arbitrary point, HK, well, then we need to shift everything over by h in the horizontal
4:36:27direction, and K in the vertical direction. So instead of the equation, x squared over
4:36:32a squared plus y squared over b squared equals one, we're gonna have to have x minus h squared
4:36:40over a squared plus y minus k squared over b squared equals one, where A is bigger than
4:36:47B, that's for an ellipse elongated in the horizontal direction. If we want an ellipse
4:36:53elongated in the vertical direction, we'll still take a bigger than B, but we'll reverse
4:36:59the roles of x and y. So this becomes x squared over b squared plus y squared over a squared
4:37:05equals one for the center at the origin. And to shifted over, we get x minus h squared
4:37:12over b squared plus y minus k squared over a squared equals one. I'll draw some pictures
4:37:19below. With some thought we can label key points. For example, this vertex must be shifted
4:37:25over by an amount a, since a is half the length of the major axis. So it'll have coordinates
4:37:31H plus a, k was this vertex will be h minus A k. This point up here should be h, k plus
4:37:41b, and this one, h k minus b, the fo cy are it, H plus c, k, and H minus C. Okay. We can
4:37:55label the points on the ellipse that's elongated vertically. Similarly, let's write the equation
4:38:02of the ellipse drawn below, and then we'll find it so cy, its major axis is in the vertical
4:38:08direction, right here. Its top vertex is at the point four, three, as bottom vertex is
4:38:16at the point for negative nine. Its center is halfway between its two vertices, which
4:38:24is right here, at the point four, negative three. Finally, I can read off the points
4:38:31at the end of the minor axis. So this one is at nine, negative three, and this one is
4:38:38at negative one, negative three. Since my lips is elongated vertically, I know I have
4:38:44an equation of the form x minus h squared over b squared plus y minus k squared over
4:38:50a squared equals one with a bigger than be my center is my point h k. And that's equal
4:38:58to four negative three, a is half the length of the major axis. So that's the vertical
4:39:05distance between this point and this point, I can just take the difference of Y coordinates
4:39:10and see that a is equal to six. B is half the length of a minor axis. So looking at
4:39:17the difference in x coordinates of these two points, I see that b is equal to five. Throwing
4:39:24that all into my equation, I get that x minus four squared over five squared plus y plus
4:39:32three squared over six squared equals one, the fo cy are going to be somewhere along
4:39:38the major axis somewhere above and below the center. To figure out how far they are, I
4:39:45need to use the fact that c squared is equal to a squared minus b squared. So in this case,
4:39:51c squared is six squared minus five squared, which is 11. Which means that C is the square
4:39:56root of 11, which is a little bit bigger than three So, the two first sigh, our distance
4:40:03of squared of 11, above and below the center, and so they'll have coordinates of four, negative
4:40:13three plus a squared of 11 and four, negative three minus the square root of 11. This video
4:40:21detailed the anatomy of ellipse, including its major axis, its minor axis, its vertices,
4:40:30and its fossa. This video is about hyperbolas. You might recall that any lips is the set
4:40:41of points x y, such that the sum of the distances between x y and two fixed points called the
4:40:52fo cy is a constant. Well, I hyperbola is a set of points x y, such that the difference
4:41:01of the distances between x y and each of two fixed points, called the first side is a constant.
4:41:17In this picture on the left, the two blue lines are the two branches of the hyperbola.
4:41:24The two red points are the focuses or foe side. So, for a point x, y on the hyperbola.
4:41:33If I look at the distance from x, y to one focus, and the distance from x, y to the other
4:41:40focus, the difference of those two distances will be the same no matter which point x y
4:41:49on the hyperbola that I choose. If I draw a line between the two fourths I, that line
4:41:58is called the transverse axis. In this left picture, we have a vertical transverse axis.
4:42:06And the hyperbola itself reminds me of a person lying down horizontally. Now in the right
4:42:12picture, our transverse axis between the two fo psi is horizontal. And the blue hyperbola
4:42:20itself kind of reminds me of a person standing up vertically. It's possible for a hyperbola
4:42:26to be oriented other ways besides horizontally and vertically, but will only consider horizontal
4:42:32and vertical orientations in this video. If you look at where the transverse axis intersects
4:42:40the blue branches of the hyperbola, you'll find these two points. those points are called
4:42:47the vertices. Notice that the vertices are the two points on the two branches that are
4:42:52closest to one another. The point halfway between the two vertices, it's also halfway
4:42:58between the two foes side is called the center of the hyperbola. The two black dotted lines
4:43:05that I've drawn that form an X are not actually part of the hyperbola. They're called the
4:43:11asymptotes. And you can think of them as guidelines for drawing the hyperbola. Because the hyperbola
4:43:17gets closer and closer to these asymptotes, but doesn't cross them. We'll talk more about
4:43:21the asymptotes later. Let's find the equation of a hyperbola with both sides at negative
4:43:27c zero, and c zero, and vertices at negative a zero and a zero. This hyperbola will have
4:43:34at center at the origin. From the distance definition of hyperbola. We know that if we
4:43:40have a point x, y on the hyperbola, then if we take the distance from that point to the
4:43:47first focus, and the distance to the second focus and subtract them, we should get a constant.
4:43:54We're always taking the positive difference here. So if we're on the right branch, we'll
4:43:59take the distance to the left focus minus the shorter distance to the right focus. And
4:44:04if we happen to be on the left branch, then we would take the longer distance to the right
4:44:12focus minus the shorter distance, the left focus, when we do the differences that way,
4:44:17we should always get the same number. And in fact, that difference should equal to A
4:44:21I'll show you why. The reason is, if I take my point to be this vertex here, then its
4:44:28distance to this focus is exactly c minus A because C is bigger than a. And this distance
4:44:37to that focus is going to be well let's see, this is a, this is a and this is c minus a.
4:44:45So that different distance will be a plus a plus c minus a. Now I'm going to do the
4:44:51difference, A plus A plus c minus a minus c minus a. That simplifies to A plus c minus
4:45:03c plus a, which is two a. Now I'll focus on a point on the right branch, I'll actually
4:45:10write a formula for its distance to this point. That's the distance formula, it's going to
4:45:15be x minus negative c squared plus y minus zero squared. That's the longer difference
4:45:24distance. Now I subtract the shorter distance, which is the distance to this focus. And they're
4:45:31the distance formula gives me x minus c squared plus y minus zero squared. That should always
4:45:37equal to A for any point on that right branch. I'll clean up the inside of the square roots
4:45:44a little bit. After that a chunk of algebra, which I won't do here allows you to simplify
4:45:51this equation to see squared minus a squared times x squared minus a squared y squared
4:45:59equals a squared times c squared minus a squared. Since this expression, c squared minus a squared
4:46:04is appearing a couple times here, I'm going to give it a new name, I'm going to let b
4:46:09squared b c squared minus a squared. Now I can rewrite my equation as b squared x squared
4:46:15minus a squared y squared equals a squared b squared. And dividing both sides by a squared
4:46:21b squared, you'll have the equation x squared over a squared minus y squared over b squared
4:46:28equals one. In this equation, notice that a represents the distance from the center
4:46:37to a vertex. The quantity C is not directly in this equation, but it's related to a and
4:46:44b by this equation, which can also be rewritten as c squared equals a squared plus b squared.
4:46:51Or if you prefer, a squared equals c squared minus b squared. I remember this one. And
4:46:57remember that C, which is the distance from the center to the focus is the largest of
4:47:02the three quantities A, B, and C. So you might be wondering, Well, what is B represent? Well,
4:47:07I'm going to draw a box, whose center is at the center of my parabola that stretches out
4:47:14in a from either direction left, right, and stretches up by B in either direction, up
4:47:21and down. It turns out that the corners of such a box will be exactly on these two asymptote
4:47:29lines. In other words, the slope of this asymptote, its rise over its run is b over a. And for
4:47:39this asymptote that's slipping downwards. Its rise over it's Ron is negative b over
4:47:45a. So I can actually write the equations of the asymptotes. For this hyperbola, centered
4:47:51at the origin as y equals B over A x and y equals negative B over A x. We did all this
4:48:04work for a standing up hyperbola with its transverse axis oriented horizontally. Let
4:48:10me rearrange things. And we'll see what changes if a hyperbola is oriented the other way.
4:48:16In this case, where we have a vertical transverse axis, the roles of x and y are completely
4:48:22switched. So our equation, which we could derive the same way, becomes y squared over
4:48:28a squared minus x squared over b squared equals one, we still have the same formulas c squared
4:48:35equals a squared plus b squared. And the two related formulas b squared equals c squared
4:48:40minus a squared, and a squared equals c squared minus b squared. A is still the distance from
4:48:46the center to the vertex. And c still represents the distance from the center to a focus. Now
4:48:53our box will extend up and down by a units and left and right by b units. So this time,
4:49:01the slope of the asymptote, the rise over the run is going to be a over b for the line
4:49:09that sloped up and negative a Overby for the line that sloped down. This means our asymptotes,
4:49:18which go through the center at the origin will be given by the equations y equals a
4:49:23over b x and y equals negative a over bx. I want to point out that in this notation,
4:49:30a is going with x in one form of the equation and y and the other form of the equation,
4:49:36but it always goes with the positive term, the term that's added and subtracted, whereas
4:49:42B goes with the negative term, the term that subtracted which term is positive there, the
4:49:49x squared or the Y squared determines the orientation of the hyperbola this way or that
4:49:55way. So far, we've been considering only hyperbola Due to the origin, and we had these two equations
4:50:02for them, if we sent her as an arbitrary point, HK, instead, we just subtract h from the x
4:50:10term and subtract k from the y term. The graphs are shifted over by age and up by K. And to
4:50:21get to the vertices, we just start at the center, and go left and right, by a, or up
4:50:32and down by a depending on which way the hyperbola is oriented. So the vertices here are at h
4:50:41plus a K, and H minus A k. Whereas over here, the vertices are at h, k plus a and h, k minus
4:50:52a. To get to the focuses, we have to go over by C, or up and down by c, where c squared
4:51:06equals a squared plus b squared. Finally, the asymptotes still have the same slopes
4:51:14as before. But this time, they have to go through the center, which is at h k instead
4:51:19of 00. Their equations are going to be given by y minus k equals B over A x minus h. And
4:51:27y minus k is negative B over A x minus h, or with the other orientation, the analogous
4:51:35thing y minus k equals a over b x minus h, and y minus k equals A over B, negative a
4:51:43over b, x minus h. We've done all the abstract theory, let's do one example with numbers.
4:51:50If we're given this hyperbola, we can read off its center as six, negative three. Since
4:51:58the x term is positive, I'm gonna let A be the square root of four, and B will be the
4:52:04square root of 25. Since the x term is positive, and the y term is negative, the hyperbola
4:52:11is going to be oriented standing up like this. With a horizontal transverse axis, the distance
4:52:20from the center to each vertex is given by the number A, which is two. So the left vertex
4:52:28will have coordinates for negative three, and the right vertex will have coordinates
4:52:33eight, negative three. To find the full side, I need to figure out c squared, which is a
4:52:41squared plus b squared, or four plus 2529. So C is the square root of 29. I'll sketch
4:52:51ballpark position of my first eye here, and the coordinates are going to be at let's see,
4:52:59the center is six, so six minus square 29, same y coordinate of negative three, and the
4:53:05six plus squared of 29 negative three. Finally, the asymptotes will form an X going through
4:53:13the center with slope given by B over A, which is five has and negative b over a negative
4:53:24five halves. The equations for those lines will be y plus three, coming from the center
4:53:32equals five halves x minus six, and y plus three equals negative five halves x minus
4:53:39six. This gives a decent rough sketch of the hyperbola. If I want to be a little more accurate,
4:53:44I could also plot some points by plugging in values of x and solving for y or vice versa.
4:53:49In this video, we gave two equations for hyperbolas. Which one you use depends on how the hyperbola
4:53:57is oriented. A has to do with the distance between the center and the vertex, the number
4:54:04c, which can be figured out from the equation c squared equals a squared plus b squared,
4:54:10is going to represent the distance from the center to the focus. And B and A together
4:54:18help determine the slope of the asymptotes. This video introduces the idea of polar coordinates.
4:54:26Polar Coordinates give an alternative way of describing the location of points on the
4:54:32plane. Instead of describing a point in terms of its x and y coordinates, those are the
4:54:39Cartesian coordinates of the point. When using polar coordinates, we instead describe the
4:54:44point in terms of radius r, and an angle theta. r is the distance of the point from the origin
4:54:53and theta is the angle that radius line makes with the positive x axis. Let's plot these
4:55:05points given in polar coordinates. So the eight here is the value of the radius, and
4:55:10the negative two thirds pi is the value of the angle theta. The negative angle means
4:55:17that I need to go clockwise from the positive x axis, instead of counterclockwise, like
4:55:22I normally would for a positive angle. So here, a negative two thirds pi means that
4:55:29I need to go to this line right here. And the eight of for the radius means I need to
4:55:36go eight lines out from the origin. So my point should be around right here. The next
4:55:42point has a radius of five and an angle of three pi. The angle of positive three pi means
4:55:51that I go counterclockwise starting at the positive x axis, here, I've gone around by
4:55:56two pi. And here, I've got an extra pi to make three pi. Now the radius of five means
4:56:03I need to go five units out from the origin. So that puts me about right here. Notice that
4:56:11I could have also labeled this point with the polar coordinates of five Pi, there's
4:56:17more than one way to assign polar coordinates to a point. The next point has an angle of
4:56:23pi over four, and a radius of negative 12. The negative radius means that I need to jump
4:56:31to the other side of the circle before I plot the point. In other words, instead of plotting
4:56:37the point at an angle of pi over four and a radius of 12, which would be about right
4:56:44here, I go to the opposite side of the circle, and plotted at the same distance from the
4:56:50origin, but 180 degrees or pi radians around the circle over here. Now I could have also
4:57:00labeled this point using a positive radius of 12. And using an angle of pi over four
4:57:07plus pi, or five pi over four. And in general, a point with polar coordinates of negative
4:57:16r theta means the same point as the point with polar coordinates r, theta plus pi, adding
4:57:23pi just makes us jump around to the opposite side of the circle. To convert between polar
4:57:33and Cartesian coordinates, it's handy to use the following equations. First, x is equal
4:57:40to r cosine theta, y is equal to r sine theta r squared is equal to x squared plus y squared,
4:57:54which means that R is plus or minus the square root of x squared plus y squared. And tangent
4:58:01theta is equal to y divided by x. Let's see where these equations come from. If we draw
4:58:09a point with coordinates, x, y, and draw lines to make a right triangle, the height of that
4:58:18triangle is y, the length of the base is x. And the high partners has length r, theta
4:58:28is the measure of this interior angle. From trig, we know that cosine theta is equal to
4:58:36adjacent over high partners. So that's x over r, which means that x is equal to r cosine
4:58:45theta. Similarly, sine theta is opposite over hypotenuse, that's y over r, which means that
4:58:54y is equal to r sine theta. That gives us the first two equations. The Pythagorean Theorem
4:59:02tells us that x squared plus y squared is equal to r squared. And that gives us the
4:59:09third equation. Finally, tangent theta is opposite over adjacent. So that's y over x,
4:59:20which is the fourth equation. To convert five, negative pi over six from polar to Cartesian
4:59:27coordinates, we just use the fact that x equals r cosine theta, and y equals r sine theta.
4:59:36So in this case, x is equal to five times cosine of negative pi over six, that's five
4:59:44times square root of three over two, and y is equal to five sine negative pi over six.
4:59:53So that's equal to negative five halves to convert negative one, negative one From Cartesian
5:00:01to polar coordinates, we know that negative one and negative one are x and y values. So
5:00:07we need to use the fact that r squared is x squared plus y squared, that is r squared
5:00:16is negative one squared plus negative one squared are two. Also, tangent theta is y
5:00:26over x, so that's negative one over negative one, or one. Now there's several values of
5:00:33r and theta that satisfy these equations are could be squared of two, or negative the square
5:00:40root of two, and theta could be Pi over four, or five pi over four. Or we could add multiples
5:00:49of two pi to either of these answers. But not all combinations of r and theta, get us
5:00:57to the right point. The point with Cartesian coordinates negative one negative one lies
5:01:04in the third quadrant. But if we use a theta value of say, pi over four and an R value
5:01:12of square root of two, that would get us to the first quadrant. So instead, we need to
5:01:17use the polar coordinates of square root of two and five pi over four. Or if we prefer,
5:01:27negative square root of two, and pi over four. We could also add any multiple of two pi to
5:01:33either of these values of theta, and get yet another way of representing the point and
5:01:38polar coordinates. This video talked about polar coordinates, and converting between
5:01:46Cartesian coordinates and polar coordinates, using some familiar equations from trig. This
5:01:55video introduces the idea of parametric equations, instead of describing a curve as y equals
5:02:02f of x, we can describe the x coordinates and y coordinates separately in terms of a
5:02:08third variable t, usually thought of as time. so we can write x as a function of t, and
5:02:18y as a separate function of t. This is especially useful as a way to describe curves that don't
5:02:27satisfy the vertical line test, and therefore can't be described traditionally as functions
5:02:33of y in terms of x. A Cartesian equation for a curve is an equation in terms of x and y
5:02:39only. parametric equations for a curve give both x and y as functions of a third variable,
5:02:47usually T. The third variable is called the parameter. That's our first example, let's
5:02:54graph the parametric equations given here on an x y coordinate axis. Well, we'll do
5:03:02this by finding x and y coordinates that correspond to the same value of t. For example, when
5:03:09t is negative two, you can calculate that x by plugging in negative two for t gives
5:03:16you five and why, when you plug in negative two for t gives you eight. Please pause the
5:03:23video for a moment and fill in some additional values of x and y. For some additional values
5:03:29of t. Your chart should look like this. And when we plot the XY pairs and connect the
5:03:35dots, we get something like this. It says this point over here corresponds to a T value
5:03:42of negative two. And this point over here corresponds to the t value of two. So if we
5:03:49think of t as time, we're traversing the curve in this direction. To find a Cartesian equation
5:03:58for this curve, we need to eliminate the variable t from these equations. One way to do this
5:04:05is to solve for t and one equation, say the first equation. So two t is equal to one minus
5:04:14x, which means that t is one half minus x over two, then we can plug that expression
5:04:21for t into the second equation and get y equals one half minus x over two squared plus four,
5:04:31which simplifies to the quadratic equation, y equals 1/4 X squared minus one half x plus
5:04:4117 fourths. Let's try some more examples. A table of values for the first example helps
5:04:50us draw the familiar graph of a circle of radius one. This should come as no surprise
5:04:58since the equations x equals cosine t and y equals sine t are familiar from trig as
5:05:05a way of describing the x&y coordinates of a point on the unit circle. Notice that when
5:05:12t equals zero, our curve lies on the positive x axis. And as t increases from zero to two
5:05:19pi, we traverse the curve once in the counterclockwise direction. A Cartesian equation for this unit
5:05:27circle is given by the equation x squared plus y squared equals one. This follows from
5:05:35the trig identity cosine squared t plus sine squared t equals one by substituting in X
5:05:43for cosine t, and y for sine t. Please pause the video for a moment to graph the second
5:05:50curve and rewrite it as a Cartesian equation. The table of values should help you see that
5:05:59the graph is again a unit circle. But this time, as t increases from zero to two pi,
5:06:06we actually traverse the circle twice in the clockwise direction, I'll draw this with a
5:06:12double arrow going clockwise. The Cartesian equation for this graph is still x squared
5:06:20plus y squared equals one. And so we found two different parameterizations. For the same
5:06:29graph on the X Y axis. Let's take a look at the third equation. There's no interval value
5:06:36specified for t here. So let's just assume that t can be any real number. Now as T ranges
5:06:43from negative infinity to infinity, our Y values, which are given by cosine t, oscillate
5:06:50between one and negative one. Our x values are always the square of our Y values. So
5:06:58the graph of this curve has to lie on the graph of x equals y squared, which is a sideways
5:07:07parabola. But a parametrically defined curve doesn't cover this whole parabola. Remember
5:07:13that y is given by cosine of t. So y can only range between negative one and one. And so
5:07:19we're only getting the portion of the parabola that I shade in here. As t varies from say,
5:07:26zero to pi, I traverse this parabola one time. And then as t goes from pi to two pi, I go
5:07:33back again in the other direction. And as T continues to increase, I traverse this parabola
5:07:40infinitely many times. The Cartesian equation for this curve is the equation x equals y
5:07:46squared with the restriction that y is between negative one and one. We've seen several examples
5:07:56where we went from parametric equations to Cartesian equations. Now let's start with
5:08:01a Cartesian equation and rewrite it as a parametric equation. In this example, y is already given
5:08:09as a function of x. So an easy way to parameterize. This curve is to just let x equal t. And then
5:08:17y is equal to the square root of t squared minus t, substituting in T for x, the domain
5:08:27restriction in terms of x just translates into a restriction in terms of t. I call this
5:08:35the copycat parameterization. Since we've successfully introduced the new variable t,
5:08:40but T just copies, whatever x does. In the second example, we could try setting x equal
5:08:47to t, then we get 25 t squared plus 36. Y squared is equal to 900. and solving for y,
5:08:57we'd have y squared equals 900 minus 25 t squared over 36. So why is plus or minus the
5:09:08square root of this quantity? This is a very awkward looking expression. In fact, why is
5:09:16that even a function of t here because of the plus and minus signs. So let's look for
5:09:20a better way to parameterize this curve. Because of the x squared and the y squared, this equation
5:09:28is a good candidate for parameterizing using sine and cosine. In fact, if we divide both
5:09:34sides of the equation by 900, we get 25x squared over 900 plus 36. Y squared over 900 is equal
5:09:45to one, which simplifies to x squared over 36 plus y squared over 25 is equal to one.
5:09:54If I rewrite this as x over six squared plus y or five squared equals one, then I can set
5:10:05x over six equal to cosine of t, and y over five equal to sine of t. And I can see that
5:10:13for any value of t, x over six and y over five will satisfy this equation, simply because
5:10:22cosine squared plus sine squared equals one. This gives me the parameterization x equals
5:10:29six cosine of t, y equals five sine of t, which is a handy way to describe any lips.
5:10:39As a final example, let's describe a general circle of radius r, and center HK. For any
5:10:51point, x, y on the circle, we know that the distance from that point x y to the center
5:11:00of the circle is equal to r. So using the distance formula, we know that the square
5:11:07root of x minus h squared plus y minus k squared has to equal our squaring both sides, this
5:11:18gives us the equation for the circle in Cartesian coordinates. So for example, if our circle
5:11:26has radius five, and has Center at the point, negative 317, then its equation would be x
5:11:35minus negative three, that's x plus three squared plus y minus 17 squared is equal to
5:11:4325. One way to find the equation of a general circle in parametric equations, is to start
5:11:52with the unit circle and work our way up. We know that the unit circle with radius one
5:11:58centered at the origin is given by the equation x equals cosine t, and y equals sine t. If
5:12:09we want a circle of radius r centered around the origin instead, then we need to expand
5:12:16everything by a factor of R. So we multiply our x and y coordinates by R. If we now want
5:12:24the center to be at HK instead of at the origin, then we need to add h to our x coordinates
5:12:35and add K to all our Y coordinates. This gives us the general equation for a circle in parametric
5:12:43equations. to match the Cartesian equation above, we can write our same example circle
5:12:49and parametric equations as x equals five cosine t minus three, y equals five sine t
5:13:01plus 17. In this video, we translated back and forth in between Cartesian equations and
5:13:11parametric equations with a special emphasis on the equations for circles. This video is
5:13:23about the difference quotient, and the average rate of change. These are topics that are
5:13:30related to the concept of derivative and calculus. For function y equals f of x, like the function
5:13:37whose graph right here, a secant line is a line that stretches between two points on
5:13:45the graph of the function. I'm going to label this x value as a and this x value as B. So
5:13:55this point here on the graph is going to have an x value of a and a y value given by f of
5:14:02a, the second point will have x value B and Y value f of b. Now the average rate of change
5:14:11for a function on the interval from a to b can be defined as the slope of the secant
5:14:23line between the two points A F of A and B f P. in symbols, that slope m is the rise
5:14:36over the run, or the change in y over the change in x, which is the difference in Y
5:14:43coordinates f of b minus F of A over the difference in x coordinates b minus a. So this is the
5:14:53average rate of change. To put this in context, if for example, f of x represents the height
5:15:01of a tree. And x represents time in years, then f of b minus f of a represents a difference
5:15:11in height, or the amount the tree grows. And B minus A represents a difference in in years,
5:15:16so a time period. So this average rate of change is the amount the tree grows in a certain
5:15:23time period. For example, if it grows 10 inches in two years, that would be 10 inches per
5:15:30two years, or five inches per year would be its average rate of change its average rate
5:15:34of growth, let's compute the average rate of change for the function f of x equals square
5:15:40root of x on the interval from one to four. So the average rate of change is f of four
5:15:49minus f of one over four minus one, well, f of four is the square root of four of one's
5:15:56a square root of one. So that's going to be two minus one over three or 1/3. Instead of
5:16:05calling these two locations on the x axis, a and b, this time, I'm going to call the
5:16:13first location, just x and the second location, x plus h. The idea is that h represents the
5:16:22horizontal distance between these two locations on the x axis. In this notation, if I want
5:16:30to label this point on the graph of y equals f of x, it'll have an x coordinate of x and
5:16:38a y coordinate of f of x. The second point will have an x coordinate of x plus h, and
5:16:44a y coordinate of f of x plus h. a difference quotient is simply the average rate of change
5:16:52using this x x plus h notation. So a different version represents the average rate of change
5:17:02of a function, f of x on the interval from x to x plus h. Equivalent way, the difference
5:17:14quotient represents the slope of the secant line for the graph of y equals f of x between
5:17:22the points with coordinates x, f of x, and x plus h, f of x plus h. Let's work out a
5:17:33formula for the difference quotient. Remember that the formula for the average rate of change
5:17:39could be written as f of b minus F of A over B minus A, where a and b are the two locations
5:17:47on the x axis. But now I'm calling instead of a I'm using x instead of B, I'm using x
5:17:54plus h. So I can rewrite this average rate of change as f of x plus h minus f of x over
5:18:03x plus h minus x. That simplifies a little bit on the denominator. Because x plus h minus
5:18:12x, I can cancel the Xs, and I get the difference quotient formula f of x plus h minus f of
5:18:21x over h. The quantity h on this nominator It looks like a single entity, but it still
5:18:28represents a difference in x values. Let's find and simplify the difference quotient
5:18:35for this function given first, or write down the general formula for the difference quotient.
5:18:41That's f of x plus h minus f of x over h. I'm going to compute f of x plus h first,
5:18:50I do this by shoving in x plus h, everywhere I see an x in the formula for the function,
5:18:58that's going to give me two times x plus h squared minus x plus h plus three. Notice
5:19:08how I use parentheses here. That's important because I need to make sure I shove in the
5:19:14entire x plus h for x. So the entire x plus h needs to be subtracted, not just the x part,
5:19:21so the parentheses are mandatory. Similarly, the parentheses here signal that the entire
5:19:27x plus h is squared as it needs to be. I mean going to go ahead and simplify a bit right
5:19:33now. I can multiply out the x plus h squared, I can go ahead and distribute the negative
5:19:42sign. So if I multiply out, I'm gonna get X squared plus X h plus h x plus h squared.
5:19:52Now I can distribute the two to get 2x squared plus 2x H plus two h x plus two h squared
5:20:02minus x minus h plus three, these two terms are actually the same, I can add them up to
5:20:08get 4x H. And I think that's as simple as I can get that part. Now I'm going to write
5:20:17out F of x plus h minus f of x. So that's going to be this thing right here, minus f
5:20:25of x. Again, I need to put the F of X formula in parentheses to make sure I subtract the
5:20:31whole thing. I'll distribute the negative. And now I noticed that a bunch of things cancel
5:20:38out. So the 2x squared and the minus 2x squared add to zero, the minus x and the x add to
5:20:45zero, and the three and the minus three add to zero. So I'm left with 4x H plus two h
5:20:55squared minus h. Finally, I'll write out the whole difference quotient by dividing everything
5:21:04by H. I can simplify this further, because notice that there's an H in every single term
5:21:10of the numerator. If I factor out this H, H times 4x, plus two h minus one divided by
5:21:18h, these two H's cancel, and I'm left with a difference quotient of 4x plus two h minus
5:21:25one. This difference quotient will become important in calculus, when we calculate a
5:21:31difference quotient for smaller and smaller values of h, eventually letting h go to zero
5:21:36and ending up with an expression that has no H's in it and represents the derivative
5:21:41or slope of the function itself. In this video, we use the formula f of b minus F of A over
5:21:49B minus A to calculate an average rate of change and the related formula f of x plus
5:21:55h minus f of x over h to calculate and simplify a difference quotient

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