SOLUTION: After transformations, the expression sin (A + B)sin (A - B) can be proved to be equal to: A) sin^2 A - sin^2 B) sin A

SOLUTION: After transformations, the expression sin (A + B)sin (A - B) can be proved to be equal to: A) sin^2 A - sin^2 B) sin A - sin B C) 2 sin A sin B D) 2 cos A cos B

Question 1198384: After transformations, the expression sin (A + B)sin (A - B) can be proved to be equal to:
A) sin^2 A - sin^2
B) sin A - sin B
C) 2 sin A sin B
D) 2 cos A cos B

Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52787) (Show Source): You can put this solution on YOUR website!
.
After transformations, the expression sin (A + B)sin (A - B) can be proved to be equal to:
A) sin^2 A - sin^2
B) sin A - sin B
C) 2 sin A sin B
D) 2 cos A cos B
~~~~~~~~~~~~~~~~~~~~~~~

Your writing in (A) is fatally defective, making the entire problem damaged and unusable.

We can not guess what is hidden there (if any).

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Comment from student: I agree with you. Can not be solved with the information given.
The problem writer must be out of his mind. Only by wild guessing, and then. Thanks anyway.

My response : very' sound judgment - a rarity in this forum.

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Answer: sin^2(A) - sin^2(B)

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Explanation:

There are at least two methods we can follow to prove that
sin(A+B)sin(A-B) = sin^2(A)-sin^2(B)
is an identity

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Method 1

Let
m = sin(A)cos(B)
n = cos(A)sin(B)

sin(A+B)sin(A-B) = [ sin(A)cos(B)+cos(A)sin(B) ] * [ sin(A)cos(B)-cos(A)sin(B) ]
sin(A+B)sin(A-B) = (m+n)(m-n)
sin(A+B)sin(A-B) = m^2 - n^2
sin(A+B)sin(A-B) = (sin(A)cos(B))^2 - (cos(A)sin(B))^2
sin(A+B)sin(A-B) = sin^2(A)cos^2(B) - cos^2(A)sin^2(B)
sin(A+B)sin(A-B) = sin^2(A)(1-sin^2(B)) - (1-sin^2(A))sin^2(B)
sin(A+B)sin(A-B) = sin^2(A)-sin^2(A)sin^2(B) - sin^2(B)+sin^2(A)sin^2(B)
sin(A+B)sin(A-B) = sin^2(A) - sin^2(B)

The relevant identities that I used were:
sin(A+B) = sin(A)cos(B)+cos(A)sin(B)
sin(A-B) = sin(A)cos(B)-cos(A)sin(B)
cos^2(A) = 1 - sin^2(A) which is from sin^2(A)+cos^2(A) = 1

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Method 2

Another identity to use is
sin(u)*sin(v) = 0.5 * [ cos(u-v) - cos(u+v) ]
which should be somewhere on your list of trig identities.

Let
u = A+B
v = A-B
So,
sin(u)*sin(v) = 0.5 * [ cos(u-v) - cos(u+v) ]
sin(A+B)*sin(A-B) = 0.5 * [ cos(A+B-(A-B)) - cos(A+B+A-B) ]
sin(A+B)*sin(A-B) = 0.5 * [ cos(2B) - cos(2A) ]
sin(A+B)*sin(A-B) = 0.5 * [ (1-2sin^2(B))-(1-2sin^2(A)) ]
sin(A+B)*sin(A-B) = 0.5 * [ 1-2sin^2(B)-1+2sin^2(A) ]
sin(A+B)*sin(A-B) = 0.5 * [ -2sin^2(B)+2sin^2(A) ]
sin(A+B)*sin(A-B) = 0.5 * [ 2sin^2(A)-2sin^2(B) ]
sin(A+B)*sin(A-B) = sin^2(A)-sin^2(B)

Edit:
Here's a handy list of trig identities
https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

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