Question

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

Answer 1

$R.H.S={\sin }^{-1}(3x-4x^3)$
Substituting $x=\sin \theta$
$\Rightarrow R.H.S={\sin }^{-1}(3\sin \theta -4{\sin }^3\theta )$
$\Rightarrow R.H.S={\sin }^{-1}\left(\sin 3\theta \right)$
$(\because \sin 3x=3\sin x-4{\sin }^{3}x)$
$x={\sin }^{-1}\theta \in [-\frac{1}{2},\frac{1}{2}]\Rightarrow \theta \in [-\frac{\pi }{6},\frac{\pi }{6}]$
$\Rightarrow 3\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$

$\Rightarrow R.H.S=3\theta$
Now,
$\because x=\sin \theta$
$\Rightarrow \theta ={\sin }^{-1}x$
$\therefore 3\theta =3{\sin }^{-1}x$
$\Rightarrow R.H.S=L.H.S$
Hence proved.