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

Answer

We know that: $\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta$
Putting $\sin \theta = x$
$ \Rightarrow \theta = {\sin ^{ - 1}}x$
$\therefore \sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta$
$\Rightarrow \sin 3\theta = 3x - 4{x^3}$
$\Rightarrow 3\theta = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)$
Putting $\theta = {\sin ^{ - 1}}x$,
$3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)$
Hence Proved.

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