Question

Evaluate the following:
i. $\sin \left({\cos }^{-1}\frac{3}{5}\right)$
ii. $\cos \left({\tan }^{-1}\frac{3}{4}\right)$
iii. $\sin (\frac{\pi }{2}-{\sin }^{-1}\left(\frac{1}{2}\right))$

• A. i. $3/5$ ii. $3/5$ iii. $\frac{\sqrt{3}}{2}$
• B. i. $3/5$ ii. $4/5$ iii. $-\frac{\sqrt{3}}{2}$
• C. i. $-4/5$ ii. $-4/5$ iii. $\frac{\sqrt{3}}{2}$
• D. i. $4/5$ ii. $4/5$ iii. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is D i. $4/5$ ii. $4/5$ iii. $\frac{\sqrt{3}}{2}$

i. Let ${\cos }^{-1}3/5=\theta$
Then, ${\cos }^{-1}3/5$ $\Rightarrow$ $\sin \theta =4/5$
$\therefore$ $\sin \left({\cos }^{-1}3/5\right)=\sin \theta =4/5$

ii. Let ${\tan }^{-1}3/4=\theta$
Then, $\tan \theta =3/5$ $\Rightarrow$ $\cos \theta =4/5$ ($\because$ ${\cos }^2\theta =\frac{1}{1+{\tan }^2\theta }$)
$\therefore$ $\cos \left({\tan }^{-1}\left(3/4\right)\right)=\cos \theta =4/5$

iii.
$\sin (\frac{\pi }{2}-{\sin }^{-1}\left(\frac{1}{2}\right))=\sin (\frac{\pi }{2}-(-\frac{\pi }{6}))=\sin \frac{2\pi }{3}=\frac{\sqrt{3}}{2}$