Answer
See below.
Work Step by Step
Use the Sum-to-Product Formula, we have:
$LHS=sin\theta[sin(3\theta)+sin(5\theta)]=sin\theta[2sin(\frac{3\theta+5\theta}{2})cos(\frac{3\theta-5\theta}{2})]
=2sin\theta sin(4\theta)cos(\theta)=sin(2\theta)sin(4\theta)$
$RHS=cos\theta[cos(3\theta)-cos(5\theta)]=cos\theta[-2sin(\frac{3\theta+5\theta}{2})sin(\frac{3\theta-5\theta}{2})]
=2cos\theta sin(4\theta)sin(\theta)= sin(2\theta)sin(4\theta)=LHS$
