Question

The real part of ${[1+\cos \left(\frac{\pi }{5}\right)+i\sin \left(\frac{\pi }{5}\right)]}^{-1}$ is

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{2}\cos \left(\frac{\pi }{10}\right)$
• D. $\frac{1}{2}\cos \left(\frac{\pi }{5}\right)$

Answer 1

The correct option is B $\frac{1}{2}$

We have,

$\begin{array}{c}\frac{1}{1+\cos \left(\frac{\pi }{5}\right)+i\sin \left(\frac{\pi }{5}\right)}\\ =\frac{1}{2{\cos }^2\frac{\pi }{10}+i2\sin \left(\frac{\pi }{10}\right)\cos \left(\frac{\pi }{10}\right)}\\ =\frac{1}{2\cos \frac{\pi }{10}\left[\cos \frac{\pi }{10}+i\sin \frac{\pi }{10}\right]}\\ =\frac{\cos \frac{\pi }{10}-i\sin \frac{\pi }{10}}{2\cos \frac{\pi }{10}}\\ Thus,Realpart=\frac{1}{2}\\ Hence,theoptionBistherequiredanswer.\end{array}$

Hence, option $B$ is the correct answer.