Question

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

• A. $-1/2$
• B. $0$
• C. $1/2$
• D. $3/2$

Answer 1

The correct option is C $1/2$

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

$=\frac{1}{2\sin \frac{2\pi }{7}}[2\sin \frac{2\pi }{7}+2\sin \frac{2\pi }{7}\cdot cos\frac{2\pi }{7}+2\sin \frac{2\pi }{7}\cdot cos\frac{4\pi }{7}+2\sin \frac{2\pi }{7}\cdot cos\frac{6\pi }{7}$

we know that $2\sin c\cos d=\sin (c+d)+\sin (c-d)$ and $\sin 2x=2\sin x\cos x$

$=\frac{1}{2\sin \frac{2\pi }{7}}[2\sin \frac{2\pi }{7}+\sin \frac{4\pi }{7}+\sin \frac{6\pi }{7}+\sin \frac{-2\pi }{7}\sin \frac{-4\pi }{7}+\sin \frac{8\pi }{7}]$
$=\frac{1}{2\sin \frac{2\pi }{7}}[2\sin \frac{2\pi }{7}+\sin \frac{4\pi }{7}+\sin \frac{6\pi }{7}-\sin \frac{2\pi }{7}-\sin \frac{4\pi }{7}+\sin \frac{8\pi }{7}]$
$=\frac{1}{2\sin \frac{2\pi }{7}}[\sin \frac{2\pi }{7}+\sin \frac{6\pi }{7}+\sin \frac{8\pi }{7}]$
$=\frac{1}{2\sin \frac{2\pi }{7}}[\sin \frac{2\pi }{7}-\sin (2\pi -\frac{6\pi }{7})+\sin \frac{8\pi }{7}]$
$=\frac{1}{2\sin \frac{2\pi }{7}}[\sin \frac{2\pi }{7}-\sin \frac{8\pi }{7}+\sin \frac{8\pi }{7}]=\frac{1}{2}$