Proving that $\sec\frac{\pi}{11}\sec\frac{2\pi}{11}\sec\frac{3\pi}{11}\sec\frac{4\pi}{11}\sec\frac{5\pi}{11}=32$ 
If $11 \gamma = \pi$ then prove that
  $
 \sec(\gamma) \sec(2\gamma) \sec(3\gamma) \sec(4\gamma) \sec(5\gamma) = 32.
$

I could not use the relation $11 \gamma = \pi$.
 A: Let
$$P=\cos\gamma\cos 2\gamma\cos 3\gamma \cos 4\gamma \cos 5\gamma$$
$$Q=\sin\gamma\sin 2\gamma\sin 3\gamma \sin 4\gamma \sin 5\gamma$$
Then, prove that $$2^5PQ=Q$$ using $$2\sin\alpha\cos\alpha=\sin(2\alpha),\quad \sin\beta=\sin(\pi-(11\gamma-\beta))$$

$$\begin{align}2^5PQ&=(2\sin\gamma\cos\gamma)(2\sin 2\gamma\cos 2\gamma)(2\sin 3\gamma\cos 3\gamma)(2\sin 4\gamma \cos 4\gamma)(2\sin 5\gamma \cos 5\gamma)\\&=\sin 2\gamma \sin 4\gamma \sin 6\gamma \sin 8\gamma \sin 10\gamma\\&=\sin 2\gamma \sin 4\gamma \sin (\pi-5\gamma) \sin(\pi-3\gamma)\sin(\pi-\gamma)\\&=\sin 2\gamma \sin 4\gamma \sin 5\gamma \sin 3\gamma \sin \gamma\\&=Q\end{align}$$

A: Like How to expand $\cos nx$ with $\cos x$?,
we can prove for positive integer $N$:
$$\cos Nx=2^{N-1}\cos^Nx+\cdots$$
If $N=2n+1$ and $\cos(2n+1)x=1, (2n+1)x=2m\pi$ where $m$ is any integer
$x=\dfrac{2m\pi}{2n+1}$ where $m\equiv0,1,2,\cdots,2n-1,2n\pmod{2n+1}$
So, the roots of $$2^{2n}\cos^{2n+1}x+\cdots-1=0$$ 
are $\cos\dfrac{2m\pi}{2n+1}$ where $m\equiv0,1,2,\cdots,2n-1,2n\pmod{2n+1}$ 
$$\implies2^{2n}\prod_{m=0}^{2n}\cos\dfrac{2m\pi}{2n+1}=(-1)^n$$
$$\implies2^{2n}\prod_{m=1}^{2n}\cos\dfrac{2m\pi}{2n+1}=(-1)^n$$
Now $\cos(\pi-A)=-\cos A,$
$$\implies2^{2n}\prod_{r=1}^n(-1)^n\cos^2\dfrac{r\pi}{2n+1}=(-1)^n$$
As for $1\le r\le n,0\le\dfrac{r\pi}{2n+1}<\dfrac\pi2\implies\cos\dfrac{r\pi}{2n+1}>0$
$$\implies\prod_{r=1}^n\cos\dfrac{r\pi}{2n+1}=\dfrac1{2^n}$$
A: $$\sec\frac{\pi}{11}\sec\frac{2\pi}{11}\sec\frac{3\pi}{11}\sec\frac{4\pi}{11}\sec\frac{5\pi}{11}=\frac{1}{\cos\frac{\pi}{11}\cos\frac{2\pi}{11}\cos\frac{3\pi}{11}\cos\frac{4\pi}{11}\cos\frac{5\pi}{11}}=$$
$$=\frac{1}{\cos\frac{\pi}{11}\cos\frac{2\pi}{11}\left(-\cos\frac{8\pi}{11}\right)\cos\frac{4\pi}{11}\left(-\cos\frac{16\pi}{11}\right)}=\frac{32\sin\frac{\pi}{11}}{32\sin\frac{\pi}{11}\cos\frac{\pi}{11}\cos\frac{2\pi}{11}\cos\frac{4\pi}{11}\cos\frac{8\pi}{11}\cos\frac{16\pi}{11}}=$$
$$=\frac{32\sin\frac{\pi}{11}}{\sin\frac{32\pi}{11}}=\frac{32\sin\frac{\pi}{11}}{\sin\left(3\pi-\frac{\pi}{11}\right)}=32$$
