Answer
$\frac{1}{2}(\sin 8x-\sin 2x)$
Work Step by Step
Use the Product-to-Sum formula $\sin u\cos v=\frac{1}{2}[\sin(u+v)+\sin(u-v)]$.
$\sin 3x\cos 5x$
$=\frac{1}{2}[\sin(3x+5x)+\sin(3x-5x)]$
$=\frac{1}{2}[\sin 8x+\sin(-2x)]$
$=\frac{1}{2}(\sin 8x-\sin 2x)$
