## Precalculus (6th Edition)

$\frac{\sin^2 x-\cos^2 x}{\csc x}=2\sin^3 x-\sin x$
Simplify the left side: $\frac{\sin^2 x-\cos^2 x}{\csc x}$ Rewrite it in terms of sine and cosine: $=\frac{\sin^2 x-\cos^2 x}{\frac{1}{\sin x}}$ Multiply top and bottom by $\sin x$: $=\frac{\sin^2 x-\cos^2 x}{\frac{1}{\sin x}}*\frac{\sin x}{\sin x}$ $=(\sin^2 x-\cos^2x)\sin x$ $=-(\cos^2x-\sin^2 x)\sin x$ Use the identity $\cos^2x-\sin^2 x=\cos 2x$: $=-\cos 2x\sin x$ Simplify the right side: $2\sin^3 x-\sin x$ Factor out $\sin x$: $=\sin x(2\sin^2 x-1)$ $=-\sin x(1-2\sin^2 x)$ Use the identity $1-2\sin^2 x=\cos 2x$: $=-\sin x\cos 2x$ $=-\cos 2x\sin x$ Since the left side and the right side both simplify to $-\cos 2x\sin x$, they are equal to each other, and the identity has been proven.