Solve it in Friday 13th - part 2

$\cos{\left(\frac{2\pi}{13} \right)}\cos{\left(\frac{6\pi}{13} \right)} + \cos{\left(\frac{2\pi}{13} \right)}\cos{\left(\frac{8\pi}{13} \right)} + \cos{\left(\frac{6\pi}{13} \right)}\cos{\left(\frac{8\pi}{13} \right)} \\ = \frac{1}{2} \left[ 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)} \right] \\ = \frac{1}{2} \left[ \cos{\left(\frac{2\pi}{13} \right)}\cos{\left(\frac{6\pi}{13} \right)} - \sin{\left(\frac{2\pi}{13} \right)}\sin{\left(\frac{6\pi}{13} \right)} \\ \quad \quad + \cos{\left(\frac{2\pi}{13} \right)}\cos{\left(\frac{6\pi}{13} \right)} + \sin{\left(\frac{2\pi}{13} \right)}\sin{\left(\frac{6\pi}{13} \right)} \\ \quad \quad \quad + \cos{\left(\frac{2\pi}{13} \right)}\cos{\left(\frac{8\pi}{13} \right)} - \sin{\left(\frac{2\pi}{13} \right)}\sin{\left(\frac{8\pi}{13} \right)} \\ \quad \quad \quad \quad + \cos{\left(\frac{2\pi}{13} \right)}\cos{\left(\frac{8\pi}{13} \right)} + \sin{\left(\frac{2\pi}{13} \right)}\sin{\left(\frac{8\pi}{13} \right)} \\ \quad \quad \quad \quad \quad + \cos{\left(\frac{6\pi}{13} \right)}\cos{\left(\frac{8\pi}{13} \right)} - \sin{\left(\frac{6\pi}{13} \right)}\sin{\left(\frac{8\pi}{13} \right)} \\ \quad \quad \quad \quad \quad \quad + \cos{\left(\frac{6\pi}{13} \right)}\cos{\left(\frac{8\pi}{13} \right)} + \sin{\left(\frac{6\pi}{13} \right)}\sin{\left(\frac{8\pi}{13} \right)} \right] \\ = \frac{1}{2} \left[ \cos{\frac{8\pi}{13}} + \cos{\frac{4\pi}{13}} + \cos{\frac{10\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{\color{#D61F06}{14}\pi}{13}} + \cos{\frac{2\pi}{13}} \right] \\ = \frac{1}{2} \left[ \cos{\frac{2\pi}{13}} + \cos{\frac{4\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} + \cos{\frac{10\pi}{13}} + \cos{\frac{\color{#D61F06}{12}\pi}{13}} \right] \\ = \frac{1}{2} \left[ \color{#3D99F6}{ - \frac{1}{2}} \right] \quad \quad \color{#3D99F6}{[\text{See Note}]} \\ = - \frac{1}{4} \\ \Rightarrow a + b = 1 + 4 = \boxed{5}$

$\color{#3D99F6}{\text{Note: }} z^{13} = e^{\frac{2\pi}{13}} = 1$ are the 13 $^{th}$ roots of unity. By Argand's diagram, we can see that: $\cos{\frac{2\pi}{13}} + \cos{\frac{4\pi}{13}} + \cos{\frac{6\pi}{13}} + \cos{\frac{8\pi}{13}} + \cos{\frac{10\pi}{13}} + \cos{\frac{12\pi}{13}} = -\frac{1}{2}$

Let $w=e^{2\pi i/13}$ , then let's use the product to sum formulas:

$\dfrac{1}{2}\left[\cos\left(\dfrac{8\pi}{13}\right)+\cos\left(\dfrac{4\pi}{13}\right)+\cos\left(\dfrac{10\pi}{13}\right)+\cos\left(\dfrac{6\pi}{13}\right)+\cos\left(\dfrac{14\pi}{13}\right)+\cos\left(\dfrac{2\pi}{13}\right)\right]$ .

But $\cos\left(\dfrac{14\pi}{13}\right)=\cos\left(\dfrac{12\pi}{13}\right)$ , so:

$\dfrac{1}{2}\left[\cos\left(\dfrac{2\pi}{13}\right)+\cos\left(\dfrac{4\pi}{13}\right)+\cos\left(\dfrac{6\pi}{13}\right)+\cos\left(\dfrac{8\pi}{13}\right)+\cos\left(\dfrac{10\pi}{13}\right)+\cos\left(\dfrac{12\pi}{13}\right)\right]$

With the fact that $\dfrac{1}{2}(w^k+w^{13-k})=\cos\left(\dfrac{2\pi k}{13}\right)$ and $w+w^2+\cdots+w^{11}+w^{12}=-1$ , which follows inmediatly from factoring $w^{13}=1$ we get:

$\dfrac{1}{2}\left(\dfrac{w+w^{12}+w^2+w^{11}+w^3+w^{10}+w^4+w^9+w^5+w^8+w^6+w^7}{2}\right) \\ \dfrac{1}{2}\left(\dfrac{-1}{2}\right)$

So, $\cos\left(\dfrac{2\pi}{13}\right)\cos\left(\dfrac{6\pi}{13}\right)+\cos\left(\dfrac{2\pi}{13}\right)\cos\left(\dfrac{8\pi}{13}\right)+\cos\left(\dfrac{6\pi}{13}\right)\cos\left(\dfrac{8\pi}{13}\right)=-\dfrac{1}{4}$

On comparing we get $\alpha=1,\beta=4$ , thus the final answer is $1+4=\boxed{5}$ .

Write a solution.