cos left(2 cos ^{-1} frac{1}{5}+sin ^{-1} fra | vedclass

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Question

(B) Let the given expression be $E = \cos \left(2 \cos ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}\right)$.We can rewrite the expression as $E = \cos \le

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the given expression be $E = \cos \left(2 \cos ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}ight)$.We can rewrite the expression as $E = \cos \left(\cos ^{-1} \frac{1}{5} + \left(\cos ^{-1} \frac{1}{5} + \sin ^{-1} \frac{1}{5}ight)ight)$.Using the identity $\cos ^{-1} x + \sin ^{-1} x = \frac{\pi}{2}$,we have $\cos ^{-1} \frac{1}{5} + \sin ^{-1} \frac{1}{5} = \frac{\pi}{2}$.Substituting this into the expression,we get $E = \cos \left(\cos ^{-1} \frac{1}{5} + \frac{\pi}{2}ight)$.Using the trigonometric identity $\cos \left(	heta + \frac{\pi}{2}ight) = -\sin 	heta$,we have $E = -\sin \left(\cos ^{-1} \frac{1}{5}ight)$.Let $	heta = \cos ^{-1} \frac{1}{5}$,then $\cos 	heta = \frac{1}{5}$.Since $\sin 	heta = \sqrt{1 - \cos^2 	heta}$,we have $\sin 	heta = \sqrt{1 - \left(\frac{1}{5}ight)^2} = \sqrt{1 - \frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{2 \sqrt{6}}{5}$.Therefore,$E = -\frac{2 \sqrt{6}}{5}$.

$\cos \left(2 \cos ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}\right)$ is equal to

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