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

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(B) Let $\cos ^{-1}\left(-\frac{3}{5}\right) = x$. Then,$\cos x = -\frac{3}{5}$. Since the range of $\cos ^{-1}$ is $[0, \pi]$,and $\cos x$ is negativ

Vedclass · Accepted Answer

$-\frac{24}{25}$

Vedclass · Answer

(B) Let $\cos ^{-1}\left(-\frac{3}{5}ight) = x$. Then,$\cos x = -\frac{3}{5}$. Since the range of $\cos ^{-1}$ is $[0, \pi]$,and $\cos x$ is negative,$x$ lies in the second quadrant. Thus,$\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \left(-\frac{3}{5}ight)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$. Now,we need to find $\sin(2x)$. Using the double angle formula,$\sin(2x) = 2 \sin x \cos x$. Substituting the values: $\sin(2x) = 2 	imes \left(\frac{4}{5}ight) 	imes \left(-\frac{3}{5}ight) = -\frac{24}{25}$.

The value of $\sin \left(2 \cos ^{-1} \left(-\frac{3}{5}\right)\right)$ is

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