Prove
2 sin 3x sin 2x = cos x − cos 5x
I'm tired and can't think straight, so this is only some stuff that will help, not the full solution.
cos 5x= 16cos(^5)x - 20cos(^3)x + 5cos x
Therefore, RHS = -16cos(^5)x + 20cos(^3)x - 4cos x
Now, LHS = either of the following (I'm tired- try and figure out yourself which is correct. I think the second one is, but I couldn't be bothered checking)
1. 2sin(^2)x.cos x(3-4sin(^2)x)
2. 2sin(^2)x.cos x(2cos(^2)x - 1 + 2cos x)
Anyway, assume both sides equal and prove true.
Divide through by cos, tidying it up a bit. Then, you can simplify -8 cos(^4)x + 10cos(^2)x -2 by changing it to:
-8 cos(^2)x.(cos(^2)x - 1) + 2cos(^2)x - 2 = -8 cos(^2)x.sin(^2)x + (-2sin(^2)x)
(same thing done on the end, to get rid of the cos)
And then you can divide through by sin(^2)x, eliminating all the sines.
I couldn't be bothered to do the rest, but it looks like it'll work out.