Prove that:

LHS $=(\sin 3 x+\sin x) \sin x+(\cos 3 x-\cos x) \cos x$

Using the identities $\sin C+\sin D=2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$ and $\cos C-\cos D=-2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$, we get

LHS $=\left(2 \sin \frac{3 x+x}{2} \times \cos \frac{3 x-x}{2} \times \sin x\right)+\left(-2 \sin \frac{3 x+x}{2} \times \sin \frac{3 x-x}{2}\right) \cos x$

$=(2 \sin 2 x \times \cos x \times \sin x)-(2 \sin 2 x \times \sin x \cos x)$

$=0=$ RHS

Hence proved.