How can I prove the fact $$\frac{1}{4-\sec^{2}\frac{2\pi}{7}} + \frac{1}{4-\sec^{2}\frac{4\pi}{7}} + \frac{1}{4-\sec^{2}\frac{6\pi}{7}} = 1.$$
When asked somebody told me to use the ideas of Chebyshev polynomial, but I haven't learnt that in school.
I tried doing this way:
Look at $y =\cos\theta + i \sin\theta$ where $\displaystyle\theta \in \Bigl\{\frac{2\pi}{7},\frac{4\pi}{7},\cdots,2\pi\Bigr\}$
Then we have
\begin{align*}
y^{7} &=1 \\ y^{7}-1 &=0 \\ (y-1) \cdot (y^{6}+y^{5}+\cdots + 1) &= 0
\end{align*}
Now the root $y=1$ corresponds to $\theta = 2\pi$, and that $$y^{6} + y^{5}+\cdots + 1 =0$$
have roots $\cos\theta + i \sin\theta$, where $\theta \in \Bigl\{\frac{2\pi}{7},\frac{4\pi}{7} ,\cdots \Bigr\}$. Looking at $y+\frac{1}{y} $ will give me the roots as $\cos\theta$ and then i can put $z=y^{2}$ to get $\cos^{2}$ as the roots and the invert to get $\sec^{2}$, but I have some problems.
Can anyone help me out with a neat solution.
Thanks.
Best Answer
Note that, for every $x$ such that this fraction exists, $$ \frac1{4-\sec^2x}=\frac14+\frac14\frac1{2\cos(2x)+1}=\frac14+\frac14\frac{\mathrm e^{2\mathrm ix}}{\mathrm e^{4\mathrm ix}+\mathrm e^{2\mathrm ix}+1}, $$ hence the identity to be proved is $$ \sum\limits_z\frac{z^2}{z^4+z^{2}+1}=1, $$ where the sum is over $z$ in $\{\omega,\omega^2,\omega^3\}$ with $\omega=\mathrm e^{2\mathrm i\pi/7}$.
If $z^7=1$ and $z\ne1$, then $\dfrac{z^2}{z^4+z^2+1}=-z^3-z^{4}$ hence the sum in the LHS is $$ -\omega^3-\omega^{4}-\omega^6-\omega^{8}-\omega^9-\omega^{12}, $$ that is, $$ -\omega^3-\omega^{4}-\omega^6-\omega-\omega^2-\omega^5=1-\sum\limits_{k=1}^7\omega^k=1. $$