Answer
$2\cos 6x\cdot\sin 3x$
Work Step by Step
Sum-to-Product:
$\displaystyle \sin A-\sin B=2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2})$
----------
$\displaystyle \frac{A+B}{2}=\frac{9x+3x}{2}=6x$
$\displaystyle \frac{A-B}{2}=\frac{9x-3x}{2}=3x$
$\sin 9x-\sin 3x=2\cos 6x\cdot\sin 3x$
