## Precalculus (6th Edition)

$4\cos 2x-4\cos 16x$
Product-to-Sum: $\displaystyle \sin A\sin B=\frac{1}{2}[\cos(A-B)-\cos(A+B)]$ $\cos(-A)=\cos A$ --------- $8\sin 7x\sin 9x=$ $=\displaystyle \frac{8}{2}[\cos(7x-9x)-\cos(7x+9x)]$ $=4[\cos(-2x)-\cos(16x)]$ $=4\cos 2x-4\cos 16x$