Question

[/tex]\tt \dfrac{9 \pi}{8} - \dfrac{9}{4} sin^{ - 1} \dfrac{1}{3} = \dfrac{9}{4} sin^{ - 1} \dfrac{2 \sqrt{2}}{3}[tex]
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Answer 1

To Prove

• $\tt \dfrac{9 \pi}{8} - \dfrac{9}{4} sin^{ - 1} \dfrac{1}{3} = \dfrac{9}{4} sin^{ - 1} \dfrac{2 \sqrt{2}}{3}$

Proof

• METHOD : 1

From L.H.S

$\implies \tt \dfrac{9 \pi}{8} - \dfrac{9}{4} sin^{ - 1} \dfrac{1}{3} \\ \\ \implies \tt \dfrac{9}{4} \left[ \dfrac{\pi}{2} - sin^{ - 1} \dfrac{1}{3} \right] \\ \\ { \underline{ \bf{ Convert \: sin \: into \: cos \: by \: the \: Pythagoras \: theorem}}} \\ \\ \implies \tt \dfrac{9}{4} \left[ \dfrac{\pi}{2} - cos^{ - 1} \: \dfrac{2 \sqrt{2} }{3} \right] \\ \\ { \boxed{ \boxed{ \bf{sin^{ - 1}x + cos^{ - 1}x = \frac{\pi}{2}}}}} \\ \\ \implies \tt \dfrac{9}{4} sin^{ - 1} \dfrac{2 \sqrt{2}}{3} \qquad { \boxed{\boxed{ \bf{sin^{ - 1}x = \frac{\pi}{2} - cos^{ - 1}x}}}}$

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• METHOD : 2

$\implies \tt \dfrac{9 \pi}{8} - \dfrac{9}{4} sin^{ - 1} \dfrac{1}{3} = \dfrac{9}{4} sin^{ - 1} \dfrac{2 \sqrt{2}}{3}$

$\implies \tt \dfrac{9}{4} sin^{ - 1} \dfrac{1}{3} + \dfrac{9}{4} sin^{ - 1} \dfrac{2 \sqrt{2}}{3} = \dfrac{9 \pi}{8}$

$\underline{ \bf{From \: L.H.S, \: take \: \frac{9}{4} \: \: as \: a \: common}}$

$\implies \tt \frac{9}{4} \left[sin^{ - 1} \dfrac{1}{3} + sin^{ - 1} \dfrac{2 \sqrt{2}}{3} \right]$

$\boxed{ \boxed{ \bf{sin^{ - 1}x + sin^{ - 1}y =sin^{ - 1}(x \sqrt{1 - y^{2}} + y \sqrt{1 - x^{2}}})}}$

$\implies \tt \frac{9}{4} \left[sin^{ - 1} \dfrac{2 \sqrt{2} }{3} \sqrt{1 - \left( \frac{1}{3} \right)^2} + \frac{1}{3} \sqrt{1 - \left(\frac{2 \sqrt{2} }{3} \right)^2 } \: \right]$

$\implies \tt \frac{9}{4} \left[sin^{ - 1} \dfrac{2 \sqrt{2} }{3} \sqrt{1 - \frac{1}{9}} + \frac{1}{3} \sqrt{1 - \frac{8}{9} } \: \right]$

$\implies \tt \frac{9}{4} \left[sin^{ - 1} \dfrac{2 \sqrt{2} }{3} \sqrt{ \frac{9 - 1}{9}} + \frac{1}{3} \sqrt{ \frac{9 - 8}{9} } \: \right]$

$\implies \tt \frac{9}{4} \left[sin^{ - 1} \dfrac{2 \sqrt{2} }{3} \times \dfrac{2 \sqrt{2} }{3} + \frac{1}{3} \times \frac{1}{3} \:\right]$

$\implies \tt \frac{9}{4} \left[sin^{ - 1} \left(\dfrac{8}{9} + \frac{1}{9} \right)\:\right]$

$\implies \tt \frac{9}{4} \left[sin^{ - 1} \left(\dfrac{9}{9} \right)\:\right]$

$\implies \tt \frac{9}{4} \left[sin^{ - 1} \left( 1 \right)\:\right]$

$\implies \tt \frac{9}{4} \left[sin^{ - 1} \left( sin \: \frac{\pi}{2} \right)\:\right]$

$\implies \tt \frac{9}{4} \times \frac{\pi}{2} = \frac{9\pi}{8}$

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Answer 2

Step-by-step explanation:

To prove: $\tt \dfrac{9 \pi}{8} - \dfrac{9}{4} sin^{ - 1} \dfrac{1}{3} = \dfrac{9}{4} sin^{ - 1} \dfrac{2 \sqrt{2}}{3}$

$\tt \dfrac{9 \pi}{8} - \dfrac{9}{4} sin^{ - 1} \dfrac{1}{3} = \dfrac{9 }{4} \left(\dfrac{\pi}{2} - sin^{ - 1} \dfrac{1}{3} \right)$

Use the identity $\tt sin^{-1}x + cos^{-1}x = \dfrac{\pi}{2}$

$\tt LHS = \dfrac{9 }{4} cos^{-1} \dfrac13$

Now use $\tt cos^{-1}x = sin^{-1} \sqrt{1-x^2}$

$\tt LHS = \dfrac{9 }{4} sin^{-1} \sqrt{1-\left(\dfrac13 \right)^2$

$\tt = \dfrac{9 }{4} sin^{-1} \sqrt{1-\dfrac19}$

$\tt = \dfrac{9 }{4} sin^{-1} \sqrt{\dfrac89}$

$\tt = \dfrac{9 }{4} sin^{-1} \dfrac{2\sqrt{2}}{3}$

$\tt = RHS$