Show that $64xy=-1$ 
Let $a=\frac{\pi}{13}$. If $$\begin{aligned} x &= \cos(2a)\cos(5a)\cos(6a)\\ y &= \cos(a)\cos(3a)\cos(9a)\end{aligned}$$ then show that $$64xy = -1$$

I'm trying to use 
$$\cos(A)\cos(B)=\frac{\cos(A+B)+\cos(A-B)}{2}$$
then use summation of $\cos(2n+1)$. I need help, please. Thank you.
 A: Hint:
Evaluate $2\sin{a}\times xy$ and recall that $\sin{u}\cos{u}=\frac{sin{2u}}{2}$

We first note that: $13a=\pi$ $\cos{9a}=\cos(13a-4a)=\cos(\pi-4a)=-\cos 4a$ and
$\cos{5a}=\cos(13a-8a)=\cos(\pi -8a)=-\cos{8a}$  also $\cos{3a}=\cos (16a-13a)=-\cos  16a$ thus:
$$2\sin{a}\times xy=\underbrace{2\sin a\cos a}_{\text{this is sin(2a)}}\cos 2a(-\cos 4a)\cos{6a}(-\cos{8a})(-\cos{16a})$$
\begin{align}
&=-\sin{2a}\cos{2a}\cos{4a}\cos{6a}\cos{8a}\cos{16a}\\
&=-\frac12\sin{4a}\cos{4a}\cos{6a}\cos{8a}\cos{16a}\\
&=-\frac14\sin{8a}\cos{8a}\cos{6a}\cos{16a}\\
&=-\frac18\sin{16a}\cos{16a}\cos{6a}\\
&=-\frac1{16}\sin{32a}\cos6a\\
&=-\frac1{16}\sin{(39a-7a)}\cos(13a-7a)\\
&=-\frac1{16}\sin7a(-\cos{7a})\\
&=\frac{1}{32}\sin{14a}\\
&=\frac{1}{32}\sin{(13a+a)}\\
&=-\frac{1}{32}\sin a.
 \end{align}
Finally:
$$2\sin a\times xy=-\frac{1}{32}\sin a$$
Divide both side by $2\sin a$ to have the needed result 
A: As $\cos(\pi-A)=-\cos A$
$\displaystyle xy=(-1)^2\prod_{r=0}^{(13-1)/2-1}\cos(2r+1)a$
We can prove something more generic $$\prod_{r=0}^{n-1}\cos\dfrac{(2r+1)\pi}{2n+1}=\dfrac{(-1)^p}{2^n}\text{ where }p=\left\lfloor\dfrac n2\right\rfloor$$
Proof: Using this , $$\cos(2n+1)x=2^{2n}\cos^{2n+1}x-\cdots+(-1)^n(2n+1)\cos x$$
If $\cos(2n+1)x=-1,(2n+1)x=(2r+1)\pi$ where $r$ is any integer
$x=\dfrac{(2r+1)\pi}{2n+1}$ where $0\le r\le2n$
So, the roots of $$2^{2n}c^{2n+1}+\cdots+(-1)^n(2n+1)c+1=0$$ are $\cos\dfrac{(2r+1)\pi}{2n+1}$ where $0\le r\le2n$
By Vieta's formulas, $$\prod_{r=0}^{2n}\cos\dfrac{(2r+1)\pi}{2n+1}=-1$$
If $2n+1=2r+1\iff n=r,\cos\dfrac{(2r+1)\pi}{2n+1}=-1$
$$\prod_{r=0, r\ne n}^{2n}\cos\dfrac{(2r+1)\pi}{2n+1}=1$$
Again as $\cos(2\pi-B)=+\cos B,$
For $r_1\ne r_2,$
$\cos\dfrac{(2r_1+1)\pi}{2n+1}=\cos\dfrac{(2r_2+1)\pi}{2n+1}\iff\dfrac{(2r_1+1)\pi}{2n+1}+\dfrac{(2r_2+1)\pi}{2n+1}=2\pi\iff r_2=2n-r_1$
$$\implies\left(\prod_{r=0}^{n-1}\cos\dfrac{(2r+1)\pi}{2n+1}\right)^2=1$$
$$\displaystyle\implies\prod_{r=0}^{n-1}\cos\dfrac{(2r+1)\pi}{2n+1}=(-1)^p$$ where $p$ is the number of $r$ such that $\cos\dfrac{(2r+1)\pi}{2n+1}<0\ \ \ \ (1)$
First of all $r<n\implies\dfrac{(2r+1)\pi}{2n+1}<\pi$
So, $(1)$ will occur if $\dfrac\pi2<\dfrac{(2r+1)\pi}{2n+1}<\pi\iff\left\lceil\dfrac{2n-1}4\right\rceil\le r\le n-1$
Observe that $r$ can have $\left\lfloor\dfrac n2\right\rfloor$ distinct values.
Here $2n+1=13\iff n=6,3\le r\le5,p=3$
