Question

${\cos }^{-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{7\pi }{5}-\sin \frac{2\pi }{5}\right)\right\}$ is equal to

• A. $\frac{23\pi }{20}$
• B. $\frac{13\pi }{20}$
• C. $\frac{3\pi }{20}$
• D. $\frac{17\pi }{20}$

Answer 1

Answer: (D)

$\frac{1}{\sqrt{2}}\left(\cos \frac{7\pi }{5}-\sin \frac{2\pi }{5}\right)=\cos \frac{\pi }{4}\cos \frac{7\pi }{5}-\sin \frac{\pi }{4}\sin \frac{2\pi }{5}=\cos \frac{\pi }{4}\cos \frac{7\pi }{5}+\sin \frac{\pi }{4}\sin \frac{7\pi }{5}$
$=\cos \left(\frac{7\pi }{5}-\frac{\pi }{4}\right)=\cos \left(\frac{23\pi }{20}\right)=\cos \left(\pi +\frac{3\pi }{20}\right)=\cos \left(\pi -\frac{3\pi }{20}\right)=\cos \frac{17\pi }{20}$
$\therefore {\cos }^{-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{7\pi }{5}-\sin \frac{2\pi }{5}\right)\right\}={\cos }^{-1}\left(\cos \left(\frac{17\pi }{20}\right)\right)=\frac{17\pi }{20}($ since $\frac{17\pi }{20}$ lies between 0 and $\pi )$