Question

How do you prove $\sin 3x=3\sin x-4{\sin }^{3}x$?

Answer 1

Proof of given relation:

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

Here we have $LHS=\sin 3x$

Therefore,

$\sin 3x=\sin \left(x+2x\right)$

$=\sin x\cos 2x+\cos x\sin 2x$ $\left[\because \sin \left(a+b\right)=\mathrm{sina}\mathrm{cosb}+\mathrm{cosa}\mathrm{sinb}\right]$

$=\sin x\left(1-2{\sin }^{2}x\right)+\cos x\left(2\cos x\sin x\right)$ $\left[\because \sin 2\theta =2\mathrm{sin\theta }\mathrm{cos\theta };\cos 2\theta =1-2{\sin }^2\theta \right]$

$=\sin x-2{\sin }^{3}x+2\sin x{\cos }^{2}x$

$=\sin x-2{\sin }^{3}x+2\sin x\left(1-{\sin }^{2}x\right)$ $\left[\because {\sin }^{2}x+{\cos }^{2}x=1\right]$

$=\sin x-2{\sin }^{3}x+2\sin x-2{\sin }^{3}x$

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

$=RHS$

Hence, $\mathbf{sin}\mathbf{}\mathbf{3}\mathbf{x}\mathbf{=}\mathbf{3}\mathbf{sin}\mathbf{}\mathbf{x}\mathbf{-}\mathbf{4}{\mathbf{sin}}^{\mathbf{3}}\mathbf{x}$ is proved.