Question

The value of $\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}$ is $=-\frac{1}{a}$ Find $a.$

Answer 1

Given, $\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}=2\cos \frac{3\pi }{7}\cos \frac{\pi }{7}+\cos \frac{6\pi }{7}$

$=2\cos \frac{3\pi }{7}\cos \frac{\pi }{7}-\cos \frac{\pi }{7}=\cos \frac{\pi }{7}(2\cos \frac{3\pi }{7}-1)$

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

$=\frac{\cos \frac{\pi }{7}(\sin \frac{6\pi }{7}-\sin \frac{3\pi }{7})}{\sin \frac{3\pi }{7}}=\frac{\cos \frac{\pi }{7}\left(2\sin \frac{3\pi }{14}\cos \frac{9\pi }{14}\right)}{2\sin \frac{3\pi }{14}\cos \frac{3\pi }{14}}$

$=\frac{\cos \frac{\pi }{7}\cos \frac{9\pi }{14}}{\cos \frac{3\pi }{14}}=\frac{1}{2}\frac{\left(\cos \frac{11\pi }{14}+\cos \frac{7\pi }{14}\right)}{\cos \frac{3\pi }{14}}$

$=\frac{1}{2}\left(\frac{-\cos \frac{3\pi }{14}+0}{\cos \frac{3\pi }{14}}\right)=-\frac{1}{2}$