Question

$1+\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}$ is equal to

• A. $\frac{-1}{2}$
• B. $0$
• C. $\frac{1}{2}$
• D. $\frac{3}{2}$

Answer 1

Answer: (C)

$\frac{1}{2}$

$1+\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}$
$=1+\frac{\cos (\frac{2\pi }{7}+\frac{6\pi }{7}).\sin (3\times \frac{2\pi }{2\times 7})}{\sin \left(\frac{2\pi }{2\times 7}\right)}[\because \cos \varphi +\cos (\alpha +\varphi )+\cos (\varphi +2\alpha )=\frac{\cos \left(\frac{\varphi +\varphi +(n+1)\alpha }{2}\right)\sin (n+1)\frac{\alpha }{2}}{\sin \frac{\alpha }{2}}]$
$=1+\frac{\cos \frac{4\pi }{7}\sin \left(\frac{3\pi }{7}\right)}{\sin \left(\frac{\pi }{7}\right)}$
$=1-\frac{\sin \frac{6\pi }{7}}{2\sin \frac{\pi }{7}}$
$=1-\frac{1}{2}=\frac{1}{2}$