Question

Given that ${\sin }^{-1}\left(\sin \frac{3\pi }{4}\right)=\frac{2\pi }{k}$, then $k=$

• A. 3
• B. 8
• C. 6
• D. 9

Answer 1

Answer: (B)

${\sin }^{-1}\left(\sin \frac{3\pi }{4}\right)={\sin }^{-1}\left\{\sin \left(\pi -\frac{\pi }{4}\right)\right\}$
$={\sin }^{-1}\left\{\sin \left(\frac{\pi }{4}\right)\right\}={\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4}=\frac{2\pi }{k}$
$\therefore k=8$