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

Let,
$\mathrm{RHS}={\sin }^{-1}\left(3\mathrm{x}-4{\mathrm{x}}^3\right)$
Put $\mathrm{x}=\sin \mathrm{\theta }$  in RHS, it follows
RHS  $={\sin }^{-1}\left(3\sin \mathrm{\theta }-4{\sin }^3\mathrm{\theta }\right)$
$\therefore \mathrm{RHS}$$={\sin }^{-1}(\sin 3\mathrm{\theta })$
Now,
$\begin{array}{l}\Rightarrow -\frac{1}{2}\le \mathrm{x}\le \frac{1}{2}\\ \Rightarrow -\frac{1}{2}\le \sin \mathrm{\theta }\le \frac{1}{2}\\ \Rightarrow -\frac{\mathrm{\pi }}{6}\le \mathrm{\theta }\le \frac{\mathrm{\pi }}{6}\\ -\frac{\mathrm{\pi }}{2}\le 3\mathrm{\theta }\le \frac{\mathrm{\pi }}{2}\end{array}$
Hence, ${\sin }^{-1}(\sin 3\mathrm{\theta })=3\mathrm{\theta }\left(\text{ as }\frac{-\mathrm{\pi }}{2}\le 3\mathrm{\theta }\le \frac{\mathrm{\pi }}{2}\right)$
$\begin{array}{l}\Rightarrow \text{ R.H.S. }=3\mathrm{\theta }\\ =3{\sin }^{-1}\mathrm{x}\\ \left[\because \mathrm{x}=\sin \mathrm{\theta }\Rightarrow \mathrm{\theta }={\sin }^{-1}\mathrm{x}\right]\end{array}$
Therefore, LHS= RHS
Therefore, $3{\sin }^{-1}\mathrm{x}={\sin }^{-1}\left(3\mathrm{x}-4{\mathrm{x}}^3\right),\mathrm{x}\in \left[-\frac{1}{2},\frac{1}{2}\right]$ has proven.