9π/8-9/4sin^1/3=9/4sin^2√2/3
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Sia ? 6 years, 2 months ago
L.H.S.= {tex}\frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\frac{1}{3} {/tex} {tex} = \frac{9}{4}\left[ {\frac{\pi }{2} - {{\sin }^{ - 1}}\frac{1}{3}} \right]{/tex}
{tex} = \frac{9}{4}{\cos ^{ - 1}}\frac{1}{3}\left[ {\because {{\sin }^{ - 1}}\frac{1}{3} + {{\cos }^{ - 1}}\frac{1}{3} = \frac{\pi }{2}} \right]{/tex}
Let
{tex}{\cos ^{ - 1}}\frac{1}{3} = \theta{/tex}
{tex}\implies\cos \theta = \frac{1}{3}{/tex}
{tex}\implies\sin \theta = \frac{{2\sqrt 2 }}{3}{/tex}
{tex}\implies\theta = {\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3}{/tex}
{tex} \Rightarrow {\cos ^{ - 1}}\frac{1}{3} = {\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3}{/tex}
{tex}\therefore L. H.S= \frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\frac{1}{3} = \frac{9}{4}{\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3} = R.H.S{/tex}
{tex}{/tex}
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