[cptn13]

[ebaines]

Sorry - not true.

sin2X = 2sinXcosX, so sin2X - sinXcosX = sinXcosX, which is not the same thing as cosX.

[harum]

sin(2x) = 2*sin(x)*cos(x), therefore:

sin(2x) - cos(x)*sin(x) = sin(x)*cos(x)

What you get is a simple equation:

sin(x)*cos(x) = cos(x) or

cos(x)*(sin(x)-1)=0

This equation holds if:

(a) cos(x)=0 => x = pi/2 + pi*n; n is any whole number, or
(b) sin(x)=1 => x = pi/2 + 2*pi*k; k is any whole number

The answer (b) is a subset of and contained in the answer (a), therefore just answer (a) would be sufficient. Hope this helps.

[Unknown008]

Yes, the answer would depend on what the question asks.

If it's a proof the identify is false, and if it's an equation then the post of harum is the way to go.