Question

The value of ${\cos }^{-1}\left[\frac{1}{\sqrt{2}}(\cos \frac{9\pi }{10}-\sin \frac{9\pi }{10})\right]$ is

• A. $\frac{17\pi }{20}$
• B. $\frac{7\pi }{20}$
• C. $\frac{7\pi }{10}$
• D. $\frac{17\pi }{10}$

Answer 1

The correct option is A $\frac{17\pi }{20}$

${\cos }^{-1}\left[\frac{1}{\sqrt{2}}(\cos \frac{9\pi }{10}-\sin \frac{9\pi }{10})\right]={\cos }^{-1}[\frac{1}{\sqrt{2}}\cos \frac{9\pi }{10}-\frac{1}{\sqrt{2}}\sin \frac{9\pi }{10}]={\cos }^{-1}\left[\cos (\frac{\pi }{4}+\frac{9\pi }{10})\right]={\cos }^{-1}\left[\cos \left(\frac{23\pi }{20}\right)\right]={\cos }^{-1}\left[\cos (\pi +\frac{3\pi }{20})\right]={\cos }^{-1}[-\cos \left(\frac{3\pi }{20}\right)]=\pi -{\cos }^{-1}\left[\cos \left(\frac{3\pi }{20}\right)\right](\because {\cos }^{-1}(-x)=\pi -{\cos }^{-1}x,x\in [-1,1\left]\right)=\pi -\frac{3\pi }{20}(\because {\cos }^{-1}(\cos x)=x,x\in [0,\pi \left]\right)=\frac{17\pi }{20}$