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Prove the following:

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

Answers (1)

D Divya Prakash Singh

Given to prove: $3\sin^{-1}x = \sin^{-1}(3x - 4x^3)$

where, $x\:\epsilon \left[-\frac{1}{2},\frac{1}{2} \right ]$ .

Take $\theta= \sin ^{-1}x$ or $x = \sin \theta$

Take R.H.S value

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

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

= $\sin^{-1}(\sin 3\theta)$ $\left ( \because 3\sin \theta - 4\sin^3 \theta \right = \sin 3 \theta )$

= $3\theta$

= $3\sin^{-1}x$ = L.H.S

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