Proving that $\sec\frac{\pi}{11}\sec\frac{2\pi}{11}\sec\frac{3\pi}{11}\sec\frac{4\pi}{11}\sec\frac{5\pi}{11}=32$

Solution 1:

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 $$2\sin\alpha\cos\alpha=\sin(2\alpha),\quad \sin\beta=\sin(\pi-(11\gamma-\beta))$$

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

Solution 2:

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 $1\le r\le n,0\le\dfrac{r\pi}{2n+1}<\dfrac\pi2\implies\cos\dfrac{r\pi}{2n+1}>0$

$$\implies\prod_{r=1}^n\cos\dfrac{r\pi}{2n+1}=\dfrac1{2^n}$$