how can I prove this identity How can I prove this identity 

$$\cos (3x/2)-\cos(x/2)=-2 \sin (x/2) \sin(x)$$

I try this but does not work with me $$\cos(3x/2) = \cos(x + x/2) = \cos (x) \cos (x/2) - \sin (x) \sin (x/2)$$ 
I do not know what to do after this step.
 A: we have $$\cos(3/2x)-\cos(x/2)=-2\sin(3/4x+x/4)\sin(x/4-x/2)$$
A: We have $$\cos(A+B) -\cos(A+B) = (\cos A\cos B -\sin A\sin B)-(\cos A\cos B +\sin A\sin B) = -2\sin A\sin B. $$ Let $A+B = C$ and $A-B = D$. Our equation then becomes $\cos C -\cos D = -2\sin (\frac{C+D}{2})\sin (\frac{C-D}{2})$. With $C=\frac{3x}{2}$ and $D=\frac{x}{2}$, we have $$ \cos(\frac{3x}{2})-\cos(\frac{x}{2}) = -2\sin (x) \sin (\frac{x}{2})$$
A: It's standard identity:-
$$\cos A-\cos B = 2\cdot\sin(\frac{B-A}{2})\cdot\sin(\frac{B+A}{2})$$
Your identity seems to be missing a minus sign.
A: I'm going to simplify the notation by letting $\frac{x}{2} = t$.  Then:
$$\cos(3x/2)-\cos(x/2)=\cos (3t) - \cos(t) = \cos (2t+t) - \cos(t) \\= \cos (2t) \cos (t) - \sin(2t)\sin(t)) -\cos t\\
= (\cos^2(t) - \sin^2 (t))\cos(t) - (2\sin(t) \cos(t) \sin(t) - \cos(t)\\
= (1-2\sin^2t)\cos t - 2\sin^2 t \cos t -\cos t\\
=\cos t - 4\sin^2 t \cos t - \cos t\\=-4\sin^2 t \cos t
= -2 (2\sin t \cos t) \sin t \\= -2\sin(2t) \sin(t) = -2\sin(x)\sin(\frac{x}{2})
$$
Your identity is missing a minus sign.
