- WELCOME TO A LESSON ON THE PRODUCT TO SUM
AND SUM TO PRODUCT TRIG IDENTITIES.
THE GOAL OF THE VIDEO IS TO USE TRIG IDENTITIES
TO REWRITE TRIGONOMETRIC EXPRESSIONS
AND ALSO TO DETERMINE FUNCTION VALUES.
SO HERE IS A LIST OF THE PRODUCT TO SUM IDENTITIES,
IF YOU TAKE A CLOSE LOOK AT THEM
YOU'LL NOTICE THEY LOOK VERY SIMILAR TO THE SUM
AND DIFFERENCE IDENTITIES FOR SINE AND COSINE
AND IN FACT THAT'S HOW THEY'RE DERIVED.
I INCLUDED THE PRODUCT TO SUM IDENTITY HERE FOR SINE.
YOU'LL NOTICE THAT TO OBTAIN IT
YOU HAVE TO TAKE THE DIFFERENCE IDENTITY FOR COSINE
AND SUBTRACT THE SUM IDENTITY.
YOU MAY WANT TO PAUSE THE VIDEO AND LOOK AT THIS MORE CLOSELY.
BUT WE'RE GOING TO GO AHEAD AND JUMP TO A COUPLE PROBLEMS.
NUMBER ONE, WE WANT TO WRITE EACH PRODUCT AS A SUM.
SO WE HAVE THE PRODUCT OF SINE -4 THETA x SINE OF 8 THETA.
SO WE HAVE SINE "A" x B
WE'RE GOING TO USE THE SECOND IDENTITY.
LET'S GO AHEAD AND SEE HOW THIS WORKS OUT.
THIS WILL EQUAL 1/2 x THE QUANTITY COSINE "A" - B
SO -4 THETA - 8 THETA - COSINE OF -4 THETA + 8 THETA
AND THAT'S PRETTY MUCH IT.
WE HAVE TO SIMPLIFY THIS AND WE NOW HAVE THE PRODUCT AS A SUM.
-4 THETA - 8 THETA WOULD BE THE COSINE -12 THETA - COSINE
-4 THETA + 8 THETA WOULD BE 4 THETA.
SO WE'VE TAKEN THIS PRODUCT AND WRITTEN IT AS A SUM.
NOW LET'S GO AHEAD AND TRY ANOTHER.
HERE WE HAVE 2 x COSINE 7T/2 x COSINE 3T/2.
NOTICE WE DO HAVE A COEFFICIENT OF 2 HERE
SO WE HAVE TO BE CAREFUL WITH THAT.
WE HAVE THE PRODUCT OF TWO COSINE FUNCTIONS
SO NOW WE'RE LOOKING AT THIS FIRST IDENTITY
WHERE ANGLE "A" IS 7T/2 AND ANGLE B IS 3T/2.
SO THIS WILL EQUAL 2 x 1/2 x THE QUANTITY COSINE OF "A" + B
+ COSINE "A" - B.
SO BE CAREFUL.
WE DID HAVE A COEFFICIENT OF 2.
THESE SIMPLIFY OUT AND WE'RE LEFT WITH COSINE OF--
THIS WOULD BE 10T/2 OR 5T AND 7T/2 - 3T/2 WOULD BE 4T/2
OR JUST 2T.
AGAIN, WE'VE REWRITTEN THIS PRODUCT AS A SUM.
HERE'S ANOTHER TYPE OF PROBLEM.
WE WANT TO FIND THE EXACT VALUE OF SINE "A" x COSINE B.
SO WE'RE LOOKING AT THIS IDENTITY NOW.
LET'S SEE HOW THIS WORKS OUT.
1/2 x THE QUANTITY OF SINE "A" + B + SINE OF THE DIFFERENCE
SO WE'RE GOING TO HAVE 1/2 AND THE SINE OF--
THIS WOULD BE 8 PI/8 OR PI.
HERE WE'RE GOING TO HAVE 7 PI/8 - 1 PI/8
THAT WOULD BE 6 PI/8 OR 3 PI/4.
SO NOW WE NEED TO FIND THESE FUNCTION VALUES.
SO THE SINE OF PI ON THE UNIT CIRCLE
THIS WOULD BE THE POINT -1 0.
REMEMBER THAT SINE IS EQUAL TO THE Y COORDINATE
ON THE UNIT CIRCLE SO THIS WOULD GIVE US 0.
NEXT, 3 PI/4 WOULD BE HERE AND OUR REFERENCE TRIANGLE
WOULD BE A 45-45 RIGHT TRIANGLE.
SO WE'D HAVE 1 1 SQUARE ROOT 2, OF COURSE THIS IS NEGATIVE.
SO THE SINE OF 3 PI/4 WOULD BE OPPOSITE OVER HYPOTHENUSE.
1/SQUARE ROOT OF 2 WHICH RATIONALIZES TO SQUARE ROOT 2/2.
SO WE HAVE 1/2 x SQUARE ROOT 2/2
WHICH EQUALS THE SQUARE ROOT OF 2/4
WHICH IS THE EXACT VALUE OF THIS PRODUCT.
SO NOW WE HAVE THE IDENTITIES THAT TAKE A SUM
AND WRITE THEM AS A PRODUCT.
I'VE INCLUDED THE DERIVATION OF ONE OF THEM
AND AGAIN, THESE ARE SIMILAR
BECAUSE THEY CAN BE VERIFIED USING THE SUM
AND DIFFERENCE FORMULAS.
SO AGAIN, I WOULD ENCOURAGE YOU TO PAUSE THE VIDEO
AND TAKE A LOOK AT THESE TO SEE WHERE THEY COME FROM.
LET'S GO AHEAD AND TRY A COUPLE PROBLEMS.
WE WANT TO WRITE THE SUM NOW AS A PRODUCT.
SO WE HAVE THE SUM OF COSINE "A" + COSINE B
SO WE'RE LOOKING AT THIS IDENTITY HERE
WHERE ANGLE "A" IS 9X AND OUR ANGLE B IS 4X.
THIS WILL BE EQUAL TO 2 x COSINE "A" + B DIVIDED BY 2.
WELL 9X + 4X IS 13X/2 x COSINE OF "A" - B DIVIDED BY 2.
WELL 9X - 4X IS 5X AND THERE IT IS.
WE'VE REWRITTEN THIS SUM AS A PRODUCT.
AGAIN, ONCE WE HAVE THESE IDENTITIES
THERE'S NOT A LOT TO IT
YOU JUST HAVE TO BE VERY CAREFUL WITH THE DETAILS.
OKAY. ON THIS LAST PROBLEM, WE WANT TO FIND THE EXACT VALUE
OF SINE 17 PI 12th - THE SINE OF 13 PI 12th.
SO WE KNOW THAT OUR ANGLE "A" WILL BE 17 PI 12th
AND ANGLE B WILL BE 13 PI 12th.
SO WE'LL BE USING THIS IDENTITY HERE.
SO THIS WILL BE EQUAL TO 2 x THE COSINE OF THE SUM
OF THE ANGLES DIVIDED BY 2.
WELL THE SUM OF THESE ANGLES WOULD BE 30 PI 12th,
WHICH WILL SIMPLIFY TO 5 PI/2.
SO WE HAVE 5 PI/2 DIVIDED BY 2
x THE SINE OF THE DIFFERENCE OF THE ANGLES DIVIDED BY 2.
THE DIFFERENCE OF THE ANGLE WOULD BE 4 PI 12s
WHICH IS EQUAL TO PI/3.
SO WE WOULD HAVE PI/3 DIVIDED BY 2.
LET'S GO AHEAD AND SIMPLIFY OUR ANGLES.
REMEMBER DIVIDING BY 2 IS THE SAME AS MULTIPLYING BY 1/2
SO THIS WOULD BE 5 PI 4th x THE SINE OF PI/6.
OKAY, WE NEED TO FIND THESE FUNCTION VALUES
AND THEN FIND THE PRODUCT.
SO AGAIN, WE DON'T HAVE OUR UNIT CIRCLE HANDY
LET'S GO AHEAD AND SKETCH THESE IN STANDARD POSITION.
5 PI/4 THERE'S PI OR 4/4 PI SO WE NEED TO GO 1/4 OF A PI MORE.
THIS PRODUCES A REFERENCE TRIANGLE
OF A 45-45-90 RIGHT TRIANGLE.
SO WE HAVE 1 1 SQUARE ROOT 2
OF COURSE BOTH THE X AND THE Y COORDINATES ARE NEGATIVE.
SO THE COSINE OF 5 PI/4 WOULD BE -1/SQUARE ROOT 2.
RATIONALIZED THAT WOULD BE -SQUARE ROOT 2/2.
NOW WE NEED TO FIND THE SINE OF PI/6.
WELL PI/6 RADIANS WOULD BE HERE WITH A REFERENCE TRIANGLE
OF A 30-60-90 SO WE HAVE 1 2 SQUARE ROOT 3,
EVERYTHING'S POSITIVE SO THE SINE OF PI/6
OPPOSITE OVER HYPOTHENUSE IS 1/2.
LET'S GO AHEAD AND SIMPLIFY THIS.
LOOKS LIKE WE HAVE -SQUARE ROOT 2 DIVIDED BY 2
AS THE EXACT VALUE OF THIS DIFFERENCE.
OKAY, THAT WILL DO IT FOR THIS VIDEO.
THANK YOU FOR WATCHING.