Answer
$${\text{We have proved that }}LHS = RHS$$
Work Step by Step
$$\eqalign{
& \sin 3\theta + \sin \theta = 2\sin 2\theta \cos \theta \cr
& \cr
& LHS = \sin 3\theta + \sin \theta {\text{ and }}RHS = 2\sin 2\theta \cos \theta \cr
& \cr
& LHS = \sin \left( {2\theta + \theta } \right) + \sin \theta \cr
& {\text{Use the identity }}\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B \cr
& LHS = \sin 2\theta \cos \theta + \cos 2\theta \sin \theta + \sin \theta \cr
& \cr
& {\text{Use the identity cos2}}\theta = 2{\cos ^2}\theta - 1 \cr
& \cr
& LHS = \sin 2\theta \cos \theta + \left( {2{{\cos }^2}\theta - 1} \right)\sin \theta + \sin \theta \cr
& LHS = \sin 2\theta \cos \theta + 2\sin \theta {\cos ^2}\theta \cr
& LHS = \sin 2\theta \cos \theta + 2\sin \theta \cos \theta \cos \theta \cr
& LHS = \sin 2\theta \cos \theta + \sin 2\theta \cos \theta \cr
& LHS = 2\sin 2\theta \cos \theta \cr
& \cr
& {\text{We have proved that }}LHS = RHS \cr} $$
