Solve this

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,

$\sin \left\{2 \cos ^{-1}\left(-\frac{3}{5}\right)\right\}=\sin (2 x)$

$=2 \sin x \cos x$

$=2 \times \frac{4}{5} \times \frac{-3}{5}$

$=-\frac{24}{25}$