# Prove $\sec^2\frac{\pi}{7}+\sec^2\frac{2\pi}{7}+\sec^2\frac{3\pi}{7}=24$ using the roots of a polynomial

I have to

prove $\sec^2\frac{\pi}{7}+\sec^2\frac{2\pi}{7}+\sec^2\frac{3\pi}{7}=24$ by using the roots of the polynomial $x^3-21x^2+35x-7=0$

I tried to factor the polynomial but it didn't work and later found it cannot factorize with rationals. and I saw some similar problems in StackExchange. But the answers are very complex to me. I cannot use Euler's complex number formula since it's not in the syllabus. I do not want the exact answer but guidance to the answer.

• Jun 4, 2018 at 1:41
• Jun 4, 2018 at 12:24

let $t=\tan(\theta)$, we have \begin{eqnarray*} \tan(7 \theta) =\frac{ 7t-35t^3+21t^5-t^7}{1-21t^2+35t^4-7t^6}. \end{eqnarray*} Set $\tan(7 \theta) =0$ then the polynomial \begin{eqnarray*} 7t-35t^3+21t^5-t^7=0 \end{eqnarray*} has roots $t=0, \tan( \pi/7), \cdots ,\tan( 6 \pi/7)$. So \begin{eqnarray*} x^3-21x^2+35x-7=0 \end{eqnarray*} has roots $x= \tan^2(\pi/7),\tan^2(2\pi/7),\tan^2(3\pi/7)$. Now let $y=x+1$.