## Precalculus (6th Edition)

$-\cot \frac{x}{2}=\frac{\sin 2x+\sin x}{\cos 2x-\cos x}$
Simplify the left side: $-\cot \frac{x}{2}$ Write cotangent as 1 divided by tangent: $=-\frac{1}{\tan \frac{x}{2}}$ Simplify: $=-\frac{1}{\frac{1-\cos x}{\sin x}}$ $=-\frac{\sin x}{1-\cos x}$ $=\frac{\sin x}{\cos x-1}$ Simplify the right side: $\frac{\sin 2x+\sin x}{\cos 2x-\cos x}$ Expand using the double-angle identities: $=\frac{2\sin x\cos x+\sin x}{2\cos^2-1-\cos x}$ $=\frac{2\sin x\cos x+\sin x}{2\cos^2-\cos x-1}$ Factor: $=\frac{\sin x(2\cos x+1)}{(2\cos x+1)(\cos x-1)}$ $=\frac{\sin x}{\cos x-1}$ Since the left side and the right side both simplify to $\frac{\sin x}{\cos x-1}$, they are equal to each other, and the identity has been proven.