Answer
$\frac{1}{2}(\sin 5x+\sin 3x)$
Work Step by Step
Use the second Product-to-Sum Formula on page 559, $\cos u\sin v=\frac{1}{2}[\sin(u+v)-\sin(u-v)]$.
$\cos x\sin 4x $
$=\frac{1}{2}[\sin(x+4x)-\sin(x-4x)]$
$=\frac{1}{2}[\sin(5x)-\sin(-3x)]$
$=\frac{1}{2}(\sin 5x+\sin 3x)$
