Prove that:

LHS $=\sin 4 x$

$=2 \sin 2 x \cos 2 x \quad(\because \sin 2 \theta=2 \sin \theta \cos \theta)$

Now, using the identities $\sin 2 \alpha=2 \sin \alpha \cos \alpha$ and $\cos 2 \alpha=\cos ^{2} \alpha-\sin ^{2} \alpha$, we get

$\mathrm{LHS}=2(2 \sin x \cos x) \cdot\left(\cos ^{2} x-\sin ^{2} x\right)$

$=4 \sin x \cos ^{3} x-4 \sin ^{3} x \cos x=$ RHS

Hence proved.