The value of sin^{-1}left(frac{2sqrt{2}}{3}ri | vedclass

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Question

(B) Let $\alpha = \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)$. Then $\sin \alpha = \frac{2\sqrt{2}}{3}$.Since $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 -

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$\frac{\pi}{2}$

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(B) Let $\alpha = \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)$. Then $\sin \alpha = \frac{2\sqrt{2}}{3}$.Since $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{8}{9} = \frac{1}{9}$,we have $\cos \alpha = \frac{1}{3}$.Thus,$\alpha = \cos^{-1}\left(\frac{1}{3}\right)$.Substituting this into the expression,we get $\cos^{-1}\left(\frac{1}{3}\right) + \sin^{-1}\left(\frac{1}{3}\right)$.Using the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$,the value is $\frac{\pi}{2}$.

The value of $\sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) + \sin^{-1}\left(\frac{1}{3}\right)$ is equal to

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