Evaluate the value of
$$\mathcal{A}=\cos \frac{2\pi}{13} + \cos \frac{6 \pi}{13} + \cos \frac{8\pi}{13}$$
Solution
It holds that $\displaystyle \sum_{n=1}^{12} \cos nt =-1$ which is a consequence of De Moivre's theorem as well as of the roots of the equation $z^{13}-1=0$. Consequently $\displaystyle \sum_{n=1}^{6} \cos nt = -\frac{1}{2}$ due to the previous note as well as of the facts $\cos 2t= \cos 24t$ and $\cos 4t = \cos 22 t$. Then for the value of $\mathcal{A}$ , if we square it, we get:
\begin{align*}
\mathcal{A}^2 &= \cos^2 \left ( \frac{2\pi}{13} \right )+ \cos^2 \left ( \frac{6 \pi}{13} \right ) + \cos^2 \left ( \frac{8 \pi}{13} \right ) + \\
& \quad +2 \cos \left ( \frac{2\pi}{13} \right )\cos \left ( \frac{6 \pi}{13} \right ) + 2 \cos \left ( \frac{2\pi}{13} \right ) \cos \left ( \frac{8\pi}{13} \right ) + 2 \cos \left ( \frac{6 \pi}{13} \right ) \cos \left ( \frac{8 \pi}{13} \right ) \\
&= \cos^2 \left ( \frac{2\pi}{13} \right )+ \cos^2 \left ( \frac{6 \pi}{13} \right ) + \cos^2 \left ( \frac{8 \pi}{13} \right ) + \\
& \quad +\cos \left ( \frac{8 \pi}{13} \right )+ \cos \left ( \frac{4 \pi}{13} \right ) + \cos \left ( \frac{14 \pi}{13} \right ) + \cos \left ( \frac{2\pi}{13} \right ) + \cos \left ( \frac{10 \pi}{13} \right ) + \cos \left ( \frac{6 \pi}{13} \right )\\
&= \cos^2 \left ( \frac{2\pi}{13} \right )+ \cos^2 \left ( \frac{6 \pi}{13} \right ) + \cos^2 \left ( \frac{8 \pi}{13} \right ) \\
& \quad + \cos \left ( \frac{8 \pi}{13} \right )+ \cos \left ( \frac{4 \pi}{13} \right ) + \cos \left ( \frac{12 \pi}{13} \right ) + \cos \left ( \frac{2\pi}{13} \right ) + \cos \left ( \frac{10 \pi}{13} \right ) + \cos \left ( \frac{6 \pi}{13} \right )\\
&= \cos^2 \left ( \frac{2\pi}{13} \right )+ \cos^2 \left ( \frac{6 \pi}{13} \right ) + \cos^2 \left ( \frac{8 \pi}{13} \right ) - \frac{1}{2}
\end{align*}
However it holds that $\cos 2a = 2 \cos^2 a -1$. Thus:
$$\mathcal{A}^2= \frac{1}{2}\left [ \cos \left ( \frac{4\pi}{13} \right ) + \cos \left ( \frac{12 \pi}{13} \right ) + \cos \left ( \frac{16 \pi}{13} \right ) +3 \right ] - \frac{1}{2}$$
Hence
$$2\mathcal{A}^2 +\mathcal{A} = \sum_{n=1}^{6} \cos nt +2 = - \frac{1}{2}+2 = \frac{3}{2}$$
Since the value of $\mathcal{A}$ is positive, we deduce that $\displaystyle \mathcal{A}= \frac{\sqrt{13}-1}{4}$.
$$\mathcal{A}=\cos \frac{2\pi}{13} + \cos \frac{6 \pi}{13} + \cos \frac{8\pi}{13}$$
Solution
It holds that $\displaystyle \sum_{n=1}^{12} \cos nt =-1$ which is a consequence of De Moivre's theorem as well as of the roots of the equation $z^{13}-1=0$. Consequently $\displaystyle \sum_{n=1}^{6} \cos nt = -\frac{1}{2}$ due to the previous note as well as of the facts $\cos 2t= \cos 24t$ and $\cos 4t = \cos 22 t$. Then for the value of $\mathcal{A}$ , if we square it, we get:
\begin{align*}
\mathcal{A}^2 &= \cos^2 \left ( \frac{2\pi}{13} \right )+ \cos^2 \left ( \frac{6 \pi}{13} \right ) + \cos^2 \left ( \frac{8 \pi}{13} \right ) + \\
& \quad +2 \cos \left ( \frac{2\pi}{13} \right )\cos \left ( \frac{6 \pi}{13} \right ) + 2 \cos \left ( \frac{2\pi}{13} \right ) \cos \left ( \frac{8\pi}{13} \right ) + 2 \cos \left ( \frac{6 \pi}{13} \right ) \cos \left ( \frac{8 \pi}{13} \right ) \\
&= \cos^2 \left ( \frac{2\pi}{13} \right )+ \cos^2 \left ( \frac{6 \pi}{13} \right ) + \cos^2 \left ( \frac{8 \pi}{13} \right ) + \\
& \quad +\cos \left ( \frac{8 \pi}{13} \right )+ \cos \left ( \frac{4 \pi}{13} \right ) + \cos \left ( \frac{14 \pi}{13} \right ) + \cos \left ( \frac{2\pi}{13} \right ) + \cos \left ( \frac{10 \pi}{13} \right ) + \cos \left ( \frac{6 \pi}{13} \right )\\
&= \cos^2 \left ( \frac{2\pi}{13} \right )+ \cos^2 \left ( \frac{6 \pi}{13} \right ) + \cos^2 \left ( \frac{8 \pi}{13} \right ) \\
& \quad + \cos \left ( \frac{8 \pi}{13} \right )+ \cos \left ( \frac{4 \pi}{13} \right ) + \cos \left ( \frac{12 \pi}{13} \right ) + \cos \left ( \frac{2\pi}{13} \right ) + \cos \left ( \frac{10 \pi}{13} \right ) + \cos \left ( \frac{6 \pi}{13} \right )\\
&= \cos^2 \left ( \frac{2\pi}{13} \right )+ \cos^2 \left ( \frac{6 \pi}{13} \right ) + \cos^2 \left ( \frac{8 \pi}{13} \right ) - \frac{1}{2}
\end{align*}
However it holds that $\cos 2a = 2 \cos^2 a -1$. Thus:
$$\mathcal{A}^2= \frac{1}{2}\left [ \cos \left ( \frac{4\pi}{13} \right ) + \cos \left ( \frac{12 \pi}{13} \right ) + \cos \left ( \frac{16 \pi}{13} \right ) +3 \right ] - \frac{1}{2}$$
Hence
$$2\mathcal{A}^2 +\mathcal{A} = \sum_{n=1}^{6} \cos nt +2 = - \frac{1}{2}+2 = \frac{3}{2}$$
Since the value of $\mathcal{A}$ is positive, we deduce that $\displaystyle \mathcal{A}= \frac{\sqrt{13}-1}{4}$.
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