Prove the following identity: cos5x - cos3x = -8sin^2x(2cos^3x - cosx)

Note that, by the sum-to-product identities:
cos(A) - cos(B) = -2sin[(A + B)/2]sin[(A - B)/2].

So, with A = 5x and B = 3x:
cos(5x) - cos(3x) = -2sin[(5x + 3x)/2]sin[(5x - 3x)/2]
==> cos(5x) - cos(3x) = -2sin(4x)sin(x).

By applying the double-angle formula for sine twice:
LHS = cos(5x) - cos(3x)
= -2sin(4x)sin(x)
= -2[2sin(2x)cos(2x)]sin(x)
= -4sin(2x)cos(2x)sin(x)
= -4[2sin(x)cos(x)]cos(2x)sin(x)
= -8sin^2(x)cos(x)cos(2x)
= -8sin^2(x)[cos(x)cos(2x)].

Then, since cos(2x) = 2cos^2(x) - 1 by the double-angle formula for cosine:
LHS = -8sin^2(x)[cos(x)cos(2x)]
= -8sin^2(x){cos(x)[2cos^2(x) - 1]}
= -8sin^2(x)[2cos^3(x) - cos(x)]
= RHS.

I hope this helps!