Question

$\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}=-\frac{1}{2}$ and $\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\cos \frac{6\pi }{7}=\frac{1}{8}$. Then the numerical value of ${cosec}^2\frac{\pi }{7}+{cosec}^2\frac{2\pi }{7}+{cosec}^2\frac{3\pi }{7}$ is

Answer 1

$cose{c}^2\frac{\pi }{7}+cose{c}^2\frac{2\pi }{7}+cose{c}^2\frac{3\pi }{7}$
$=\frac{1}{{\sin }^2\frac{\pi }{7}}+\frac{1}{{\sin }^2\frac{2\pi }{7}}+\frac{1}{{\sin }^2\frac{3\pi }{7}}$
$=\frac{2}{1-\cos \frac{2\pi }{7}}+\frac{2}{1-\cos \frac{4\pi }{7}}+\frac{2}{1-\cos \frac{6\pi }{7}}$
Now denominator (after taking LCM)
$=(1-\cos \frac{2\pi }{7})(1-\cos \frac{4\pi }{7})(1-\cos \frac{6\pi }{7})$
$=1-(\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7})+(\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}+\cos \frac{4\pi }{7}\cos \frac{6\pi }{7}+\cos \frac{6\pi }{7}+\cos \frac{2\pi }{7})$
$-\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\cos \frac{6\pi }{7}$
The values of the second and last terms are known.
The third term can be written as
$\frac{1}{2}[\cos \frac{6\pi }{7}+\cos \frac{2\pi }{7}+\cos \frac{10\pi }{7}+\cos \frac{2\pi }{7}+\cos \frac{8\pi }{7}+\cos \frac{4\pi }{7}]$
$=\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}$
$=-\frac{1}{2}$
The denominator can be written as
$1-(-\frac{1}{2})-\frac{1}{2}-\frac{1}{8}$
$=\frac{7}{8}$
The numerator can be written as
$=2[3-2[\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}]+\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}+\cos \frac{4\pi }{7}\cos \frac{6\pi }{7}+\cos \frac{6\pi }{7}\cos \frac{2\pi }{7}]$
$=2[3-2(-\frac{1}{2})-\frac{1}{2}]$
$=7$.
Hence, the given expression $=\frac{7}{7/8}=8$