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)

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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