Trig proof: sum of squared cosecants

[Greg]

Hi! I've tried a couple of approaches with this: converting to complex exponential form and using standard trigonometric identities but have been unable to solve. I suspect DeMoivre's formula applies but I don't see how.Prove \(\displaystyle \csc^2\left(\dfrac{\pi}{7}\right)+\csc^2\left(\dfrac{2\pi}{7}\right)+\csc^2\left(\dfrac{4\pi}{7}\right)=8\)

 

[kaliprasad]

Hi! I've tried a couple of approaches with this: converting to complex exponential form and using standard trigonometric identities but have been unable to solve. I suspect DeMoivre's formula applies but I don't see how.Prove \(\displaystyle \csc^2\left(\dfrac{\pi}{7}\right)+\csc^2\left(\dfrac{2\pi}{7}\right)+\csc^2\left(\dfrac{4\pi}{7}\right)=8\)

you should be able to solve with the following approach ( if there is a problem I shall provide solution tomorrow)

using $\tan\, 7x = 0$ you can find the solution $\tan\, nx$ to be root of $t^6-21t^4+35t^2-7=0$
form the eqaution whose roots are $\frac{1}{t^2}$
you should have the roots $\cot^2 nx$
from this should know the sum of $\cot^2 nx$
add 1 to each term and get the result

 

[Greg]

using $\tan\, 7x = 0$ you can find the solution $\tan\, nx$ to be root of $t^6-21t^4+35t^2-7=0$

I don't understand how to derive this result.

 

[kaliprasad]

I don't understand how to derive this result.

from the series expansion of $\tan (2n+1) x$ we have

$\tan 7x = \dfrac{\tan ^7 x -21\tan ^ 5x + 35tan^3 x - 7\ tan\, x }{7\tan^6 x - 35 \tan ^4 x + 21 \tan ^2 x -1}$

as $\tan7x = 0$ for $x = \dfrac{n\pi}{7}$ n = 0 to 7 so

$\tan\,nx$ is root of$t^7 -21t^5 + 35t^3 - 7t= 0 $

as for n = 1 to 6 $\tan\, nx$ is not zero so
n = 1 to 6 $\tan\, nx$ is root of $t^6 -21t^4 + 35t^2 - 7= 0$
$\cot \, nx$ is root of $(\dfrac{1}{t})^6 -21(\dfrac{1}{t})^4 + 35(\dfrac{1}{t})^2 - 7= 0 $

or $7t^6-35t^4 +21t^2 -1=0$
so $\cot^2 nx$ is root of equation $7t^3 - 35 t^2 + 21 t - 1$ = 0
now $\cot^2 x = \cot^2 6x$
$\cot^2 2x = \cot^2 5 x$
$\cot^2 3x = \cot^2 4x$
so the roots are $\cot^2 x,\cot^2 2x,\cot^2 4x$
so we get sum of roots
$\cot^2 x + \cot^2 2x + \cot^2 4x = 5$ and adding 1 to each term on LHS you get the result

edit: I remembered that I have solved a similar problem at http://mathhelpboards.com/challenge-questions-puzzles-28/cotangent-sum-16172.html
and you can put n = 3 to get the result.