Question

Prove that:
$\sin 4x=4\sin x{\cos }^{3}x-4\cos x{\sin }^{3}x$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\mathrm{sin\; 4}x \\ =2\sin 2x\cos 2x\left(\because \sin 2\mathrm{\theta }=2\mathrm{sin\theta }\mathrm{cos\theta }\right) \end{gathered}$$

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

$$\begin{gathered} \mathrm{LHS}=2(2\sin x\cos x).({\cos }^{2}x-{\sin }^{2}x) \\ =4\sin x{\cos }^{3}x-4{\sin }^{3}x\cos x=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$