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

Answer 1

(d) $-\frac{24}{25}$

Let ${\cos }^{-1}\left(-\frac{3}{5}\right)=x,0\le x\le \mathrm{\pi }$
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,
$$\begin{gathered} \sin \left\{2{\cos }^{-1}\left(-\frac{3}{5}\right)\right\}=\sin \left(2x\right) \\ =2\sin x\cos x \\ =2\times \frac{4}{5}\times \frac{-3}{5} \\ =-\frac{24}{25} \end{gathered}$$