# How do you evaluate?

## cos^2((pi)/13)+cos^2((2pi)/13)+cos^2((3pi)/13)+cos^2((4pi)/13)+cos^2((5pi)/13)=?

Jan 16, 2018

${\cos}^{2} \left(\frac{\pi}{13}\right) + {\cos}^{2} \left(\frac{2 \pi}{13}\right) + {\cos}^{2} \left(\frac{3 \pi}{13}\right) + {\cos}^{2} \left(\frac{4 \pi}{13}\right) + {\cos}^{2} \left(\frac{5 \pi}{13}\right) + {\cos}^{2} \left(\frac{6 \pi}{13}\right) = \frac{11}{4}$

#### Explanation:

The $13$th roots of $1$ are:

$\cos \left(\frac{2 n \pi}{13}\right) + i \sin \left(\frac{2 n \pi}{13}\right)$

for $n = - 6 , - 5 , - 4 , - 3 , - 2 , - 1 , 0 , 1 , 2 , 3 , 4 , 5 , 6$

Noting that:

$\left\{\begin{matrix}\cos \left(- \theta\right) = \cos \theta \\ \sin \left(- \theta\right) = - \sin \theta\end{matrix}\right.$

we can take the roots in pairs to find:

${x}^{13} - 1 = \left(x - 1\right) \left({x}^{2} - 2 \cos \left(\frac{2 \pi}{13}\right) x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{4 \pi}{13}\right) x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{6 \pi}{13}\right) x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{8 \pi}{13}\right) x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{10 \pi}{13}\right) x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{12 \pi}{13}\right) x + 1\right)$

$\textcolor{w h i t e}{{x}^{13} - 1} = \left(x - 1\right) \left({x}^{12} + {x}^{11} + {x}^{10} + \ldots + x + 1\right)$

Equating the coefficients of ${x}^{11}$, we find:

$- 2 \left(\cos \left(\frac{2 \pi}{13}\right) + \cos \left(\frac{4 \pi}{13}\right) + \cos \left(\frac{6 \pi}{13}\right) + \cos \left(\frac{8 \pi}{13}\right) + \cos \left(\frac{10 \pi}{13}\right) + \cos \left(\frac{12 \pi}{13}\right)\right) = 1$

So:

$\cos \left(\frac{2 \pi}{13}\right) + \cos \left(\frac{4 \pi}{13}\right) + \cos \left(\frac{6 \pi}{13}\right) + \cos \left(\frac{8 \pi}{13}\right) + \cos \left(\frac{10 \pi}{13}\right) + \cos \left(\frac{12 \pi}{13}\right) = - \frac{1}{2}$

Now:

$\cos 2 \theta = 2 {\cos}^{2} \theta - 1$

Hence:

${\cos}^{2} \theta = \frac{1}{2} \left(\cos 2 \theta + 1\right)$

So we find:

${\cos}^{2} \left(\frac{\pi}{13}\right) + {\cos}^{2} \left(\frac{2 \pi}{13}\right) + {\cos}^{2} \left(\frac{3 \pi}{13}\right) + {\cos}^{2} \left(\frac{4 \pi}{13}\right) + {\cos}^{2} \left(\frac{5 \pi}{13}\right) + {\cos}^{2} \left(\frac{6 \pi}{13}\right)$

$= \frac{1}{2} \left(\cos \left(\frac{2 \pi}{13}\right) + \cos \left(\frac{4 \pi}{13}\right) + \cos \left(\frac{6 \pi}{13}\right) + \cos \left(\frac{8 \pi}{13}\right) + \cos \left(\frac{10 \pi}{13}\right) + \cos \left(\frac{12 \pi}{13}\right)\right) + 3$

$= - \frac{1}{4} + 3$

$= \frac{11}{4}$