#### Answer

$\frac{\sin 2x}{2\sin x}=\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}$

#### Work Step by Step

Let's simplify the left side:
$\frac{\sin 2x}{2\sin x}$
Use the identity $\sin 2x=2\sin x\cos x$:
$=\frac{2\sin x\cos x}{2\sin x}$
$=\cos x$
Simplify the right side:
$=\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}$
Use the identity $\cos 2A=\cos^2 A-\sin^2 A$, where $A=\frac{x}{2}$:
$=\cos (2*\frac{x}{2})$
$=\cos x$
Since the left side and the right side are both equal to $\cos x$, they are equal to each other, and the identity has been proven.