Question:
$\sin \left\{2 \cos ^{-1}\left(\frac{-3}{5}\right)\right\}$ is equal to
(a) $\frac{6}{25}$
(b) $\frac{24}{25}$
(c) $\frac{4}{5}$
(d) $-\frac{24}{25}$
Solution:
(d) $-\frac{24}{25}$
Let $\cos ^{-1}\left(-\frac{3}{5}\right)=x, 0 \leq x \leq \pi$
Then, $\cos x=-\frac{3}{5}$
$\therefore \sin x=\sqrt{1-\cos ^{2} x}=\sqrt{1-\left(-\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$
Now,
$\sin \left\{2 \cos ^{-1}\left(-\frac{3}{5}\right)\right\}=\sin (2 x)$
$=2 \sin x \cos x$
$=2 \times \frac{4}{5} \times \frac{-3}{5}$
$=-\frac{24}{25}$
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