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

Answer: (D)

$\sin \left\{2{\cos }^{-1}\left(-\frac{3}{5}\right)\right\}$
$=2\sin \left\{{\cos }^{-1}\left(-\frac{3}{5}\right)\right\}\cos \left\{{\cos }^{-1}\left(-\frac{3}{5}\right)\right\}$
$=2\sin \left\{\pi -{\cos }^{-1}\frac{3}{5}\right\}\times \left(-\frac{3}{5}\right)$
$=\frac{-6}{5}\sin \left\{{\cos }^{-1}\frac{3}{5}\right\}=\frac{-6}{5}\sin \left({\sin }^{-1}\sqrt{1-\frac{9}{25}}\right)$
$=\frac{-6}{5}\times \frac{4}{5}=\frac{-24}{25}$