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Class-12-science»Maths

Show that 9 pi /8 -9/4 sin^-1 1/3 = 9/4 sin^-1 2 root 2/3

\frac{9}{4} \sin^{- 1} (\frac{2 \sqrt{2}}{3}) + \frac{9}{4} \sin^{- 1} (\frac{1}{3}) = \frac{9}{4} [\sin^{- 1} (\frac{2 \sqrt{2}}{3}) + \sin^{- 1} (\frac{1}{3})] = \frac{9}{4} [\sin^{- 1} (\frac{2 \sqrt{2}}{3} \times \sqrt{1 - \frac{1}{9}} + \frac{1}{3} \sqrt{1 - \frac{8}{9}})] = \frac{9}{4} [\sin^{- 1} (\frac{8}{9} + \frac{1}{9})] = \frac{9}{4} \sin^{- 1} (1) = \frac{9}{4} \times \frac{π}{2} = \frac{9 π}{8} So, \frac{9 π}{8} - \frac{9}{4} \sin^{- 1} (\frac{1}{3}) = \frac{9}{4} \sin^{- 1} (\frac{2 \sqrt{2}}{3})

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