To prove that cos(3x)cos(2x) = sin(3x)sin(2x), we can first expand both sides using the trigonometric identities:

cos(3x)cos(2x) = (cos(3x + 2x) + cos(3x - 2x))/2
= (cos(5x) + cos(x))/2

sin(3x)sin(2x) = (cos(3x - 2x) - cos(3x + 2x))/2
= (cos(x) - cos(5x))/2

Now we can compare these two expressions:

(cos(5x) + cos(x))/2 = (cos(x) - cos(5x))/2

Simplifying by multiplying both sides by 2, we get:

cos(5x) + cos(x) = cos(x) - cos(5x)

Rearranging terms, we get:

2cos(5x) = 0

This is true since cosine values repeat every 2π, and cos(0) = cos(2π) = 1. Therefore, the identity cos(3x)cos(2x) = sin(3x)sin(2x) is valid.