Question

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

Answer 1

Let $x=\sin \theta$.

Since $x\in [-\frac{1}{2},\frac{1}{2}]$ that means, $\theta \in [-\frac{\pi }{6},\frac{\pi }{6}]$

$\Rightarrow 3{\sin }^{-1}x=3\theta$ and $3\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$ $...\left(1\right)$

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

$3x-4x^3\in [-1,1]$ since $x\in [-\frac{1}{2},\frac{1}{2}]$

$\Rightarrow {\sin }^{-1}(3x-4x^3)={\sin }^{-1}\left(\sin 3\theta \right)=3\theta$ $...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get,

$3{\sin }^{-1}x={\sin }^{-1}(3x-4x^3)$

Hence Proved.