#### Answer

$4\cos 2x-4\cos 16x$

#### Work Step by Step

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