#### Answer

$3\cos x-3\cos 9x$

#### Work Step by Step

Product-to-Sum:
$\displaystyle \sin A\sin B=\frac{1}{2}[\cos(A-B)-\cos(A+B)]$
$\cos(-x)=\cos x$
---------
$6\sin 4x\sin 5x=$
$=\displaystyle \frac{6}{2}[\cos(4x-5x)-\cos(4x+5x)]$
$=3[\cos(-x)-\cos(9x)]$
$=3\cos x-3\cos 9x$