Question

Evaluate the following:
i. ${\sin }^{-1}\left(\frac{2\pi }{4}\right)$
ii. ${\cos }^{-1}\left(\cos \frac{7\pi }{6}\right)$
iii. ${\tan }^{-1}\left(\tan \frac{2\pi }{3}\right)$
iv. $\cos ({\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)+\frac{\pi }{6})$

• A. i. $-\frac{\pi }{3}$ ii. $-\frac{5\pi }{6}$ iii. $\frac{\pi }{3}$ iv. $1$
• B. i. $\frac{2\pi }{3}$ ii. $\frac{\pi }{6}$ iii. $-\frac{2\pi }{3}$ iv. $1$
• C. i. $-\frac{2\pi }{3}$ ii. $-\frac{\pi }{6}$ iii. $\frac{2\pi }{3}$ iv. $-1$
• D. i. $\frac{\pi }{3}$ ii. $\frac{5\pi }{6}$ iii. $-\frac{\pi }{3}$ iv. $-1$

Answer 1

Answer: (D)

i. $\frac{\pi }{3}$ ii. $\frac{5\pi }{6}$ iii. $-\frac{\pi }{3}$ iv. $-1$

i. ${\sin }^{-1}\left(\frac{2\pi }{4}\right)\ne \frac{2\pi }{3}$, as $\frac{2\pi }{3}$ does not lie between $-\frac{\pi }{2}$ and $\frac{\pi }{2}$ Now, ${\sin }^{-1}\left(\sin \frac{2\pi }{3}\right)={\sin }^{-1}\left(\sin (\pi -\frac{2\pi }{3})\right)={\sin }^{-1}\left(\sin \frac{\pi }{3}\right)=\frac{\pi }{3}$
ii. ${\cos }^{-1}\left(\cos \frac{7\pi }{6}\right)\ne \frac{7\pi }{6}$, as
$\frac{7\pi }{6}$ does not lie between 0 and $\pi$
Now, ${\cos }^{-1}\left(\cos \frac{7\pi }{6}\right)={\cos }^{-1}\left(\cos (2\pi -\frac{5\pi }{6})\right)={\cos }^{-1}\left(\cos \frac{5\pi }{6}\right)=\frac{5\pi }{6}$
iii. ${\tan }^{-1}\left(\tan \frac{2\pi }{3}\right)\ne \frac{2\pi }{3}$, because $\frac{2\pi }{3}$ does not lie between $-\frac{\pi }{2}$ and $\frac{\pi }{2}$
Now, ${\tan }^{-1}\left(\tan \frac{2\pi }{3}\right)$ $={\tan }^{-1}\left(\tan (\pi -\frac{\pi }{3})\right)={\tan }^{-1}(-\tan \frac{\pi }{3})={\tan }^{-1}\left(\tan (-\frac{\pi }{3})\right)=-\frac{\pi }{3}$
iv. $\cos ({\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)+\frac{\pi }{6})=\cos (\frac{5\pi }{6}+\frac{\pi }{6})=\cos \left(\pi \right)=-1$