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!