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Prove that 9 pi / 8 - 9 / 4 sin inverse 1 / 3 = 9 / 4 sin inverse 2 root 2 / 3

Prove that

    12. \frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\frac{1}{3} = \frac{9}{4}\sin^{-1}\frac{2\sqrt2}{3}

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We have to solve the given equation:

\frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\frac{1}{3} = \frac{9}{4}\sin^{-1}\frac{2\sqrt2}{3} 

Take \frac{9}{4} as common in L.H.S,

=\frac{9}{4}\left [ \frac{\pi}{2}- \sin^{-1}\frac{1}{3} \right ]

or =\frac{9}{4}\left [ \cos^{-1}\frac{1}{3} \right ]    from      \left [ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \right ]

Now, assume,

  \left [ \cos^{-1}\frac{1}{3} \right ] = y

Then,

\cos y = \frac{1}{3} \Rightarrow \sin y = \sqrt{1-(\frac{1}{3})^2} = \frac{2.\sqrt2}{3}

Therefore we have now,

y = \sin^{-1} \frac{2.\sqrt2}{3}

So we have L.H.S then = \frac{9}{4}\sin^{-1} \frac{2.\sqrt2}{3}

That is equal to R.H.S.

Hence proved.

 

 

 

 

 

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