Question

${\sin }^{-1}\left(117/125\right)$ is equal to

• A. $2\alpha$
• B. $3\alpha$
• C. $\pi /2-2\alpha$
• D. $\pi -3\alpha$

Answer 1

Answer: (B), (D)

$3\alpha$
$\pi -3\alpha$

$\alpha ={\cos }^{-1}\left(\frac{4}{5}\right)\Rightarrow \cos \alpha =\frac{4}{5}\Rightarrow \sin \alpha =\frac{3}{5}\Rightarrow \tan \alpha =\frac{3}{4}\beta ={\tan }^{-1}\frac{2}{3}\Rightarrow \tan \beta =\frac{2}{3}\Rightarrow \sin \beta =\frac{2}{\sqrt{3}}\Rightarrow \cos \beta =\frac{3}{\sqrt{13}}$

Option A: $\sin 2\alpha =2\sin \alpha \cos \alpha =2(\frac{3}{5}\times \frac{4}{5})=\frac{24}{25}\ne \frac{117}{125}$

Option B: $\sin 3\alpha =3\sin \alpha -4{\sin }^3\alpha =3\left(\frac{3}{5}\right)-4{\left(\frac{3}{5}\right)}^3=\frac{9}{5}-\frac{108}{125}=\frac{117}{125}$

$\therefore {\sin }^{-1}\left(\frac{117}{125}\right)=3\alpha$

Hence, option B is correct.

Option C: $\sin (\frac{\pi }{2}-2\alpha )=\cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha$

$={\left(\frac{4}{5}\right)}^2-{\left(\frac{3}{5}\right)}^2=\frac{7}{25}\ne \frac{117}{125}$

Option D: $\sin (\pi -3\alpha )=\sin [\frac{\pi }{2}+(\frac{\pi }{2}-3\alpha \left)\right]=\cos (\frac{\pi }{2}-3\alpha )$

$=\sin 3\alpha$

$=\frac{117}{125}$

Hence, option D is also correct.