Show that $64xy=-1$

Question

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.

Answer 1

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

Answer 2

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 $2n+1=13\iff n=6,3\le r\le5,p=3$