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