The value of cos left(2 cos ^{-1} x+sin ^{-1} | vedclass

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Question

(D) Given the expression $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$.We can rewrite the argument as $\cos \left(\cos ^{-1} x + (\cos ^{-1} x + \si

Vedclass · Accepted Answer

$-\frac{2 \sqrt{6}}{5}$

Vedclass · Answer

(D) Given the expression $\cos \left(2 \cos ^{-1} x+\sin ^{-1} xight)$.We can rewrite the argument as $\cos \left(\cos ^{-1} x + (\cos ^{-1} x + \sin ^{-1} x)ight)$.Using the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,the expression becomes $\cos \left(\cos ^{-1} x + \frac{\pi}{2}ight)$.Using the trigonometric identity $\cos \left(\frac{\pi}{2} + 	hetaight) = -\sin 	heta$,we get $-\sin \left(\cos ^{-1} xight)$.Since $\sin \left(\cos ^{-1} xight) = \sqrt{1-x^2}$,the expression simplifies to $-\sqrt{1-x^2}$.Substituting $x = \frac{1}{5}$:$-\sqrt{1-\left(\frac{1}{5}ight)^2} = -\sqrt{1-\frac{1}{25}} = -\sqrt{\frac{24}{25}} = -\frac{2 \sqrt{6}}{5}$.

The value of $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ at $x=\frac{1}{5}$,where $0 \leq \cos ^{-1} x \leq \pi$ and $-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$,is

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