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

(N/A) To prove $3 \sin ^{-1} x = \sin ^{-1}(3 x - 4 x^{3})$,where $x \in [-\frac{1}{2}, \frac{1}{2}]$.
Let $x = \sin \theta$. Then,$\theta = \sin ^{-1} x$.
Since $x \in [-\frac{1}{2}, \frac{1}{2}]$,we have $\sin \theta \in [-\frac{1}{2}, \frac{1}{2}]$,which implies $\theta \in [-\frac{\pi}{6}, \frac{\pi}{6}]$.
Multiplying by $3$,we get $3\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
Now,consider the $R$.$H$.$S$.:
$\sin ^{-1}(3 x - 4 x^{3}) = \sin ^{-1}(3 \sin \theta - 4 \sin ^{3} \theta)$
Using the trigonometric identity $\sin 3\theta = 3 \sin \theta - 4 \sin ^{3} \theta$,we get:
$= \sin ^{-1}(\sin 3 \theta)$
Since $3\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,we can use the property $\sin ^{-1}(\sin \alpha) = \alpha$ for $\alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
$= 3 \theta$
$= 3 \sin ^{-1} x = \text{L.H.S.}$
Hence,the identity is proved.