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

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

Solution:

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

Let $x=\sin \theta$. Then, $\sin ^{-1} x=\theta$.

We have,

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

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

$=3 \theta$

$=3 \sin ^{-1} x$

$=L . H . S .$

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