Find the closed form of $\cos \frac{2 \pi}{13} + \cos \frac{6 \pi}{13} + \cos \frac{8 \pi}{13}$

Question

From the problem 1 from this link.

I was trying to prove that

$$\cos \dfrac{2 \pi}{13} + \cos \dfrac{6 \pi}{13} + \cos \dfrac{8 \pi}{13} = \dfrac{\sqrt{13} - 1}{4} $$

Please provide me some hint.

Answer 1

Let $C_{k} = \cos \left(\dfrac{k \pi}{13} \right)$

Lemma:

$$\color{blue}{C_{2} + C_{4} + C_{6} + C_{8} + C_{10} + C_{12} = -\dfrac{1}{2} }$$

Proof :

Observe that $C_{2k} = \cos \left(\dfrac{2k \pi}{13} \right) = \Re (e^{(2k i\pi)/13})$

So, $$\color{red}{C_{2} + C_{4} + C_{6} + C_{8} + C_{10} + C_{12} = \Re \left(\sum_{k=1}^{6} e^{(2k i\pi)/13} \right)}$$

It is a geometric series. So, evaluating it and finding the real part proves our lemma.

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Properties of $C_{k}$ :

• $C_{2k} = 2 C_{k}^2 - 1$.

• $2C_{p} C_{q} = C_{p+q} + C_{p-q}$.

• $C_{13+k} = C_{13-k}$.

The above properties can be proved easily.

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Now we are required to find the value of $\cos \dfrac{2 \pi}{13} + \cos \dfrac{6 \pi}{13} + \cos \dfrac{8 \pi}{13}$, which is equal to $C_{2} + C_{6} + C_{8}$.

Let $x = C_{2} + C_{6} + C_{8}$. Squaring on both sides yields

$$x^2 = C_{2}^2 + C_{6}^2 + C_{8}^2 + 2C_{2}C_{6} + 2 C_{2}C_{8} + 2 C_{6}C_{8} \\ x^2 = \dfrac{C_{4} + 1}{2} + \dfrac{C_{12} + 1}{2} + \dfrac{C_{16} + 1}{2} + C_{8} + C_{4} + C_{10} + C_{6} + C_{14} + C_{2} \\ 2x^2 = C_{4} + C_{12} + C_{16} + 2C_{8} + 2C_{4} + 2C_{10} + 2C_{6} + 2C_{14} + 2C_{2} + 3 \\ 2x^2 = 3C_{4} + (C_{12} + 2C_{14}) + (C_{16} + 2 C_{10}) + 2(C_2 + C_6 + C_8) +3$$

Now, using the third property that $C_{13+k} = C_{13-k}$, we get $C_{12} = C_{14}$ and $C_{16} = C_{10}$. Then,

$$2x^2 = 3C_{4} + 3C_{12} + 3C_{10} + 2(C_2 + C_6 + C_8) +3 \\ 2x^2 = 3(C_{4} + C_{10} + C_{12}) + 2x +3$$

From our lemma,

$$C_{2} + C_{4} + C_{6} + C_{8} + C_{10} + C_{12}= - \dfrac{1}{2} \\ C_{4} + C_{10} + C_{12} = - \dfrac{1}{2} - (C_2 + C_6 + C_8) = - \dfrac{1}{2} - x$$

So,

$$2x^2 = 3(C_{4} + C_{10} + C_{12}) + 2x +3 = 3(- \dfrac{1}{2} - x) + 2x +3 \\ 4x^2 = -2x+3 \\ 4x^2 + 2x -3 =0 \implies x = \dfrac{-1 \pm \sqrt{13}}{4}$$

We are getting two values of $x$. Now, observe that $\cos \dfrac{6 \pi}{13} > 0$. And also $\cos \dfrac{2 \pi}{13} + \cos \dfrac{8 \pi}{13} = \cos \dfrac{2 \pi}{13} - \cos \dfrac{5 \pi}{13}$. But $\cos$ function is strictly decreasing i.e.

$$\dfrac{2 \pi}{13} < \dfrac{5 \pi}{13} \implies \cos \dfrac{2 \pi}{13} > \cos \dfrac{5 \pi}{13} \\ \implies \cos \dfrac{2 \pi}{13} - \cos \dfrac{5 \pi}{13} > 0 \\ \cos \dfrac{2 \pi}{13} + \cos \dfrac{6 \pi}{13} - \cos \dfrac{5 \pi}{13} > 0$$

So, $\cos \dfrac{2 \pi}{13} + \cos \dfrac{6 \pi}{13} + \cos \dfrac{8 \pi}{13} > 0$. This implies that the required value is $\dfrac{\sqrt{13} - 1}{4}$.

Therefore,

$$\color{green}{\cos \dfrac{2 \pi}{13} + \cos \dfrac{6 \pi}{13} + \cos \dfrac{8 \pi}{13} = \dfrac{\sqrt{13} - 1}{4}}$$