## Precalculus: Mathematics for Calculus, 7th Edition

$\frac{1}{2}(\cos 4x-\cos 6x)$
Using the product-to sum formula on page 559, $\sin u\sin v=\frac{1}{2}[\cos (u-v)-\cos(u+v)]$: $\sin x\sin 5x$ $=\frac{1}{2}[\cos(x-5x)-\cos(x+5x)]$ $=\frac{1}{2}[\cos(-4x)-\cos 6x]$ $=\frac{1}{2}(\cos 4x-\cos 6x)$