## Precalculus (6th Edition)

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