Question

Solve : $cos\frac{2\pi }{7}+cos\frac{4\pi }{7}+cos\frac{6\pi }{7}=$

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

Answer 1

The correct option is D $-\frac{1}{2}$

Given,

$\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}$

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

multiplying each term by $\sin \left(\frac{\pi }{7}\right)$ and converting from product to sum form, we get,

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

$=\frac{\sin (-\frac{\pi }{7})+\sin \left(3\frac{\pi }{7}\right)+\sin (-3\frac{\pi }{7})+\sin \left(5\frac{\pi }{7}\right)+\sin (-5\frac{\pi }{7})+\sin \left(7\frac{\pi }{7}\right)}{2\sin \left(\frac{\pi }{7}\right)}$

upon further simplification, we get,

$=\frac{\sin (-\frac{\pi }{7})+\sin \pi }{2\sin \left(\frac{\pi }{7}\right)}$

$=\frac{\sin (-\frac{\pi }{7})+0}{2\sin \left(\frac{\pi }{7}\right)}$

$=-\frac{1}{2}$