Trigonometric sum

Prove that

$\cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7}=\frac{1}{2}$

Solution

Consider the tridiagonal matrix $A =\begin{pmatrix} 1&1&0\\ 1&0&1\\ 0&1&0 \end{pmatrix}$ . Its eigenvalues are $\displaystyle 2 \cos \frac{\pi}{7} \; , \; 2 \cos \frac{3\pi}{7} \; , \; 2 \cos \frac{5 \pi}{7}$ . Hence,

$\Tr\left ( A \right ) = 1 = 2 \left ( \cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7} \right )$

and the result follows.

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Another solution.

The points $\displaystyle \left (\cos \frac {(2n+1)\pi}{7},\sin\frac {(2n+1)\pi}{7} \right ) \; , \; n \in \{0, 1,2, \dots , 6 \}$ are vertices of a regular septagon. It follows from the rotational symmetry of the polygon that:

$\begin{align*} \sum_{i=0}^{6} \cos \frac{\left ( 2i+1 \right )\pi}{7} =0 &\Leftrightarrow 2\left ( \cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7} \right ) + \cos \pi =0 \\ &\Leftrightarrow \cos \frac{\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{5\pi}{7} = \frac{1}{2} \end{align*}$

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