Sum and Difference Identities for Sine — Transcript
Full transcript
- 0:11- WELCOME TO A VIDEO
- 0:12ON THE SUM AND DIFFERENCE IDENTITIES FOR SINE.
- 0:14THE GOAL OF THE VIDEO
- 0:15IS TO USE THE SUM AND DIFFERENCE IDENTITIES
- 0:17TO DETERMINE FUNCTION VALUES.
- 0:20SO THE SINE OF A SUM OR DIFFERENCE
- 0:22ARE EQUAL TO THE FOLLOWING.
- 0:24THE SINE OF THE QUANTITY A + B
- 0:25IS EQUAL TO SINE A x COSINE B + COSINE A x SINE B.
- 0:32AND IF WE HAVE A DIFFERENCE,
- 0:33OR THE SINE OF THE QUANTITY A - B,
- 0:36THAT'S EQUAL TO SINE A x COSINE B - COSINE A x SINE B.
- 0:42NOW, WE CAN COMBINE THESE TWO IDENTITIES
- 0:44INTO A SINGLE IDENTITY IF WE WRITE IT LIKE THIS.
- 0:47NOTICE WE HAVE A +/- SIGN HERE AND A +/- SIGN HERE.
- 0:51SO IF WE HAVE A SUM OF TWO ANGLES,
- 0:52WE HAVE A SUM OF THESE PRODUCTS.
- 0:54IF WE HAVE A DIFFERENCE OF TWO ANGLES,
- 0:56WE HAVE THE DIFFERENCE OF THESE TWO PRODUCTS.
- 0:59NOW, I DID INCLUDE A SLIDE
- 1:01THAT SHOWS THE PROOF OF THE SINE SUM IDENTITY.
- 1:04WE DON'T HAVE TIME TO GO THROUGH ALL OF THIS NOW,
- 1:07BUT IF YOU READ THROUGH THIS, IT'S PRETTY STRAIGHTFORWARD,
- 1:10AND IT WILL VERIFY THE VALIDITY
- 1:11OF THE SUM AND DIFFERENCE IDENTITY FOR SINE.
- 1:16LET'S GO AHEAD AND TAKE A LOOK AT SOME PROBLEMS.
- 1:19WE WANT TO DETERMINE THE SINE OF A - B
- 1:22GIVEN SINE A EQUALS 4/5 IN THE SECOND QUADRANT,
- 1:26AND COSINE B = -5/13 IN THE THIRD QUADRANT.
- 1:31LET'S GO AHEAD AND SKETCH THESE ANGLES
- 1:34IN STANDARD POSITION.
- 1:36SO FOR ANGLE WHOSE TERMINAL SIDE
- 1:38IS IN THE SECOND QUADRANT,
- 1:41IT MAY LOOK SOMETHING LIKE THIS,
- 1:45AND SO WE'LL CALL THIS REFERENCE ANGLE A PRIME.
- 1:48SINCE THE SINE OF THIS ANGLE IS EQUAL TO 4/5,
- 1:51WE KNOW THE OPPOSITE SIDE IS EQUAL TO 4,
- 1:53THE HYPOTENUSE IS EQUAL TO 5.
- 1:55THIS IS A 3, 4, 5 RIGHT TRIANGLE,
- 1:57SO THIS WOULD BE EQUAL TO -3
- 1:59SINCE WE ARE IN THE SECOND QUADRANT.
- 2:02NEXT, COSINE B IS EQUAL TO -5/13 IN THE THIRD QUADRANT.
- 2:10SO THIS WOULD BE ANGLE B IN STANDARD POSITION.
- 2:13SO IF WE DRAW OUR REFERENCE TRIANGLE,
- 2:16IT MIGHT LOOK SOMETHING LIKE THIS,
- 2:18SO WE'LL CALL OUR REFERENCE ANGLE B PRIME
- 2:21AND THE COSINE OF THIS ANGLE IS EQUAL TO -5/13,
- 2:27AND THIS IS 5, 12, 13 RIGHT TRIANGLE,
- 2:31AND SO THIS WOULD BE -12.
- 2:34SO NOW LET'S GO BACK TO OUR IDENTITY,
- 2:38AND WE SHOULD BE ABLE TO FIND ALL OF THESE VALUES NOW.
- 2:41WE HAVE THE SINE OF ANGLE A - B = SINE A x COSINE B.
- 2:51WELL, SINE A IS GIVEN AS 4/5,
- 2:56COSINE B IS ALSO GIVEN AS - 5/13.
- 3:00SINCE WE'RE USING A DIFFERENCE HERE
- 3:01WE'LL HAVE A SUBTRACTION SIGN HERE.
- 3:04NOW, THE COSINE OF ANGLE A
- 3:05WE CAN USE THIS REFERENCE TRIANGLE HERE
- 3:08ADJACENT OVER HYPOTENUSE, THAT'S -3/5.
- 3:13AND FOR THE SINE OF ANGLE B
- 3:14WE'LL USE THIS REFERENCE TRIANGLE OPPOSITE
- 3:17OVER HYPOTENUSE OR -12/13.
- 3:22NOW AGAIN, NOTICE WE HAVE A COMMON DENOMINATOR
- 3:25OF 5 x 13 OR 65.
- 3:29OUR FIRST NUMERATOR IS -20 - THIS WILL BE POSITIVE 36,
- 3:38SO THE SINE OF A - B WILL EQUAL -56/65.
- 3:45AND THAT'S WHAT WE WERE LOOKING FOR IN THIS PROBLEM.
- 3:49LET'S TAKE A LOOK AT ANOTHER EXAMPLE NOW.
- 3:51WE WANT TO DETERMINE THE EXACT VALUE OF SINE 150 DEGREES.
- 3:56150 DEGREES IS NOT ONE OF THOSE NICE REFERENCE ANGLES,
- 4:01BUT WE CAN USE A SUM OR DIFFERENCE OF REFERENCE ANGLES
- 4:04TO OBTAIN 105 DEGREES.
- 4:06105 DEGREES IS EQUAL TO 60 DEGREES + 45 DEGREES,
- 4:14BOTH OF WHICH ARE REFERENCE ANGLES
- 4:17THAT WE CAN USE IN THIS IDENTITY
- 4:18TO DETERMINE THE EXACT VALUE OF SINE 105 DEGREES.
- 4:22SO LET'S GO AHEAD AND SET IT UP.
- 4:31SO SINCE THE SINE OF 105 DEGREES
- 4:32EQUALS THE SINE OF 60 + 45 DEGREES,
- 4:36OUR VALUE FOR A WILL BE 60 DEGREES
- 4:38AND OUR VALUE FOR B WILL BE 45 DEGREES.
- 4:43SO THIS WILL EQUAL THE SINE OF A x THE COSINE OF
- 4:57+ COSINE OF A x THE SINE OF B.
- 5:08AND WE'LL USE A UNIT CIRCLE TO FIND THESE VALUES
- 5:11AND THEN FIND THEIR PRODUCTS AND THEN FIND THE SUM.
- 5:14SO THE SINE OF 60 DEGREES,
- 5:16REMEMBER SINE IS EQUAL TO Y ON THE UNIT CIRCLE.
- 5:20SQUARE ROOT 3 OVER 2 TIMES THE COSINE OF 45 DEGREES
- 5:26EQUAL TO THE X COORDINATE SQUARE ROOT 2 OVER 2
- 5:30PLUS COSINE OF 60 WHICH IS EQUAL TO 1/2,
- 5:37AND THE SINE OF 45 SQUARE ROOT 2 OVER 2.
- 5:43SO WHAT WE HAVE FOUND IS THE SINE OF 105 DEGREES
- 5:48IS EQUAL TO THE SUM OF THESE PRODUCTS.
- 5:54NOTICE OUR DENOMINATOR IS 4 IN BOTH CASES.
- 5:57HERE THE NUMERATOR IS THE SQUARE ROOT OF 6
- 6:00+ THE SQUARE ROOT OF 2.
- 6:06OKAY. I THINK I HAVE ONE MORE EXAMPLE.
- 6:11NOW WE HAVE AN ANGLE THAT'S IN RADIANS,
- 6:12SO IT MIGHT BE HELPFUL TO CONVERT THIS TO DEGREES
- 6:15SO THAT WE CAN DETERMINE WHICH REFERENCE ANGLES TO USE.
- 6:24THIS EQUALS -15 DEGREES.
- 6:27SO WHAT WE CAN DO IS WE CAN USE 30 DEGREES - 45 DEGREES
- 6:36AND THAT WILL GIVE US THE -15 DEGREES THAT WE NEED.
- 6:39BUT LET'S BE CONSISTENT AND LET'S USE RADIANS.
- 6:43SO 30 DEGREES IS THE SAME AS PI OVER 6 RADIANS,
- 6:48AND 45 DEGREES IS THE SAME AS PI OVER 4 RADIANS.
- 6:56SO WE'LL USE THIS FOR ANGLE A AND THIS FOR ANGLE B.
- 6:59LET'S SET IT UP.
- 7:03SO WE'LL HAVE THE SINE OF PI OVER 6 - PI OVER 4,
- 7:07WHICH AGAIN WILL GIVE US THE -PI OVER 12.
- 7:12SO WE'LL HAVE THE SINE OF A,
- 7:13WHICH IS THE SINE OF PI OVER 6 x THE COSINE OF ANGLE B
- 7:20WHICH IS PI OVER 4 - COSINE OF ANGLE A OR COSINE OF PI OVER 6
- 7:30x THE SINE OF B WHICH IS PI OVER 4.
- 7:39USING OUR UNIT CIRCLE,
- 7:40THE SINE OF PI OVER 6 WOULD BE 1/2,
- 7:46COSINE OF PI OVER 4 IS SQUARE ROOT 2 OVER 2,
- 7:52COSINE OF PI OVER 6 WOULD BE SQUARE ROOT 3 OVER 2,
- 7:57AND THE SINE OF PI OVER 4,
- 7:59AGAIN, IS SQUARE ROOT 2 OVER 2.
- 8:01OKAY. SO NOW WE HAVE THE VALUE OF SINE -PI OVER 12.
- 8:06NOTICE WE HAVE A COMMON DENOMINATOR OF 4.
- 8:10THIS NUMERATOR WOULD BE THE SQUARE ROOT OF 2 -
- 8:14AND THIS IS THE SQUARE ROOT OF 6,
- 8:16AND WE HAVE THE EXACT VALUE OF SINE -PI OVER 12.
- 8:23OKAY. THAT'S PRETTY MUCH ALL WE HAVE TIME FOR.
- 8:25I HOPE YOU FOUND THIS VIDEO HELPFUL.
- 8:28THANK YOU FOR WATCHING.
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