Question

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

Answer 1

Proof of given relation:

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

Here we have $LHS=\sin 3\theta$

Therefore,

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

$=\sin \theta \cos 2\theta +\cos \theta \sin 2\theta$ $\left[\because sin\left(a+b\right)=sinacosb+cosasinb\right]$

$=\sin \theta \left(1-2{\sin }^2\theta \right)+\cos \theta \left(2\cos \theta \sin \theta \right)$ $\left[\because sin2\theta =2sin\theta cos\theta ;cos2\theta =1-2si{n}^2\theta \right]$

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

$=\sin \theta -2{\sin }^3\theta +2\sin \theta \left(1-{\sin }^2\theta \right)$ $\left[\because si{n}^{2}x+co{s}^{2}x=1\right]$

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

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

$=RHS$

Hence, $\mathit{s}\mathit{i}\mathit{n}\mathbf{3}\mathit{\theta }\mathbf{=}\mathbf{3}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{-}\mathbf{4}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{3}}\mathit{\theta }$ is proved.