Question

Prove: $3{\sin }^{-1}x={\sin }^{-1}(3x-{4x}^3),x\in [-\frac{1}{2},\frac{1}{2}]$

Answer 1

Let $x=\sin \theta$
$\Rightarrow {\sin }^{-1}x=\theta$

Thus we have,

$R.H.S.={\sin }^{-1}(3x-{4x}^3)$
$={\sin }^{-1}(3\sin \theta -4{\sin }^3\theta )$

$={\sin }^{-1}\left(\sin 3\theta \right)$ [$\because \sin 3A=3\sin A-4{\sin }^{3}A$]
$=3\theta$ [$\because {\sin }^{-1}\left(\sin A\right)=A$]
$=3{\sin }^{-1}x$
$=\text{L.H.S.}$
Therefore, $3{\sin }^{-1}x={\sin }^{-1}(3x-{4x}^3)$