cos left[2 sin ^{-1} frac{3}{4} + cos ^{-1} | vedclass

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Question

(C) Let $ x = \sin ^{-1} \frac{3}{4} $. Then $\sin x = \frac{3}{4}$.We know that $\sin ^{-1} \frac{3}{4} + \cos ^{-1} \frac{3}{4} = \frac{\pi}{2}$.The

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$ -\frac{3}{4} $

Vedclass · Answer

(C) Let $ x = \sin ^{-1} \frac{3}{4} $. Then $\sin x = \frac{3}{4}$.We know that $\sin ^{-1} \frac{3}{4} + \cos ^{-1} \frac{3}{4} = \frac{\pi}{2}$.Therefore,the expression becomes $\cos \left[\sin ^{-1} \frac{3}{4} + (\sin ^{-1} \frac{3}{4} + \cos ^{-1} \frac{3}{4})ight]$.$= \cos \left[\sin ^{-1} \frac{3}{4} + \frac{\pi}{2}ight]$.Using the identity $\cos \left(\frac{\pi}{2} + 	hetaight) = -\sin 	heta$,we get:$= -\sin \left(\sin ^{-1} \frac{3}{4}ight)$.$= -\frac{3}{4}$.

$ \cos \left[2 \sin ^{-1} \frac{3}{4} + \cos ^{-1} \frac{3}{4}\right] $

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