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