Question

The value of $\cos \left(\frac{2\pi }{7}\right)+\cos \left(\frac{4\pi }{7}\right)+\cos \left(\frac{6\pi }{7}\right)$ is equal to :

• A. $-1$
• B. $-\frac{1}{2}$
• C. $-\frac{1}{3}$
• D. $-\frac{1}{4}$

Answer 1

Answer: (B)

$\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}$
$=\frac{\sin \left(3\times \frac{\pi }{7}\right)}{\sin \frac{\pi }{7}}\times \cos \left(\frac{\frac{2\pi }{7}+\frac{6\pi }{7}}{2}\right)$
$=\frac{2\sin \left(\frac{3\pi }{7}\right)}{2\sin \frac{\pi }{7}}\times \cos \left(\frac{4\pi }{7}\right)$
$=\frac{\sin \left(\frac{7\pi }{7}\right)+\sin \left(\frac{-\pi }{7}\right)}{2\sin \frac{\pi }{7}}$
$=\frac{-\sin \frac{\pi }{7}}{2\sin \frac{\pi }{7}}$
$=-\frac{1}{2}$