## Precalculus: Mathematics for Calculus, 7th Edition

$\frac{1}{2}(\sin 5x-\sin x)$
Using the product-to-sum formulas on page 559, $\sin u\cos v=\frac{1}{2}[\sin(u+v)+\sin(u-v)]$: $\sin 2x\cos 3x$ =$\frac{1}{2}[\sin(2x+3x)+\sin(2x-3x)]$ =$\frac{1}{2}[\sin 5x+\sin (-x)]$ =$\frac{1}{2}(\sin 5x-\sin x)$