Question

The values of $\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}=$

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

Answer 1

Answer: (C)

$\frac{-1}{2}$

$Formulaused$
$2\sin A\cos B=\mathrm{Sin}(A+B)+\sin (A-B)$
$\mathrm{Sin}(-\theta )=-\sin \theta$
$Given$
$\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}$
$multiplydivideby2\sin \frac{\pi }{7}$
$\frac{1}{2\sin \frac{\pi }{7}}[2\sin \frac{\pi }{7}\cos \frac{2\pi }{7}+2\sin \frac{\pi }{7}\cos \frac{4\pi }{7}+2\sin \frac{\pi }{7}\cos \frac{6\pi }{7}]$
$\Rightarrow \frac{1}{2\sin \frac{\pi }{7}}[\sin \frac{3\pi }{7}+\sin (-\frac{\pi }{7})+\sin \frac{5\pi }{7}+\sin (-\frac{3\pi }{7})+\sin \frac{7\pi }{7}+\sin (-\frac{5\pi }{7})]$
$\Rightarrow \frac{1}{2\sin \frac{\pi }{7}}[\sin \frac{3\pi }{7}-\sin \left(\frac{\pi }{7}\right)+\sin \frac{5\pi }{7}-\sin \left(\frac{3\pi }{7}\right)+\sin \frac{\pi }{}-\sin \left(\frac{5\pi }{7}\right)]$
$\Rightarrow \frac{1}{2\sin \frac{\pi }{7}}[-\sin \left(\frac{\pi }{7}\right)+\sin \pi ][\sin \pi =0]$
$\Rightarrow \frac{-\sin \frac{\pi }{7}}{2\sin \frac{\pi }{7}}$
$\Rightarrow \frac{-1}{2}$