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

 

 

 

 
 
 
 
 

Answers (1)

RHS = \sin^{-1}\left ( 3x-4x^{3} \right )
putting x= \sin \theta  in RHS we get
RHS = \sin^{-1}\left ( 3\sin \theta -4\sin ^{3}\theta \right )= \sin^{-1}\left ( \sin 3\theta \right )
Now , \frac{-1}{2}\leq x\leq \frac{1}{2}
\Rightarrow \frac{-1}{2}\leq \sin \theta \leq \frac{1}{2}
\Rightarrow \frac{-\pi }{6}\leq \theta \leq \frac{\pi }{6}
\Rightarrow \frac{-\pi }{2}\leq 3\theta \leq \frac{\pi }{2}
Hence \sin^{-1}\left ( \sin 3\theta \right )= 3\theta \left [ as\: \frac{-\pi }{2} \leq 3\theta \leq \frac{\pi }{2}\right ]
RHS = 3\theta = 3\sin^{-1}\left ( x \right )\left [ \because x= \sin \theta \Rightarrow \theta = \sin^{-1}x \right ]
                  = LHS Hence proved.

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