Going backwards?!

Q:  Express sin9x – sin7x as a product containing only sines and/or cosines:

A. 2sinx cos8x
B. -2sinx cos8x
C. 2sin8x cos x
D. -2sin8x cosx

A:  This problem is almost exactly like this problem, only it is going backwards (and has sines instead of cosines… but it is the same concept).

Use the following identity:

cos(a)sin(b) = [sin(a + b) – sin(a – b)] / 2

Multiply both sides by 2:

2cos(a)sin(b) = sin(a + b) – sin(a – b)

Notice that our problem looks like the right side a little?  We have:  sin9x – sin7x

Watch this magic!

sin9x – sin7x = sin(8x + x) – sin(8x – x)

See how our problem now looks like that identity?

sin(a + b) – sin(a – b) = sin(8x + x) – sin(8x – x) = 2cos(8x)sin(x)

This is the same as answer A!

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