I would like to know how to prove that
$$\prod_{r=1}^{(n-1)/2} 2\cos\frac{\pi r}{n}=1~~~~~\text{where $n$ is odd}$$
I have tried induction and using trigonometric identities to attempt to transform the product into something more helpful, but nothing has worked.
It may be more helpful to think of the product as $$\prod_{r=1}^{n} 2\cos\frac{\pi r}{2n+1}=1$$
Many thansk for your help.
@Somos's tip is useful. In the second parameterization, write the product as$$\frac{\prod_{r=1}^n\sin\frac{2\pi r}{2n+1}}{\prod_{r=1}^n\sin\frac{\pi r}{2n+1}},$$then note the numerator's sines have the same values as the denominator's, but in another order. In particular, replacing $2r$ with $2n+1-2r$ when $2r\ge n+1$ produces a permutation of $\{1,\,\cdots,\,n\}$. For example, if $n=2$ the numerator is$$\sin\frac{2\pi}{5}\sin\frac{4\pi}{5}=\sin\frac{2\pi}{5}\sin\frac{\pi}{5}.$$
Using this
$$\cos(2n+1)x=2^{2n}\cos^{2n+1}x+\cdots+(-1)^n\cos x$$
If $\cos(2n+1)x=-1=\cos\pi\implies(2n+1)x=(2r+1)\pi$
$\implies x=\dfrac{(2r+1)\pi}{2n+1}; 0\le r\le2n$
So, the roots of $$2^{2n}c^{2n+1}+\cdots+(-1)^nc+1=0$$ are $\cos\dfrac{(2r+1)\pi}{2n+1}; 0\le r\le2n$
Using Vieta's Formulas $$\prod_{r=0}^{2n}\cos\dfrac{(2r+1)\pi}{2n+1}=-\dfrac1{2^{2n}}$$
Exclude $r=n$
$$\prod_{r=0,\ne n}^{2n}\cos\dfrac{(2r+1)\pi}{2n+1}=\dfrac1{2^{2n}}$$
Now if $r_1+r_2=2n+1,\cos\dfrac{(2r_1+1)\pi}{2n+1}=\cos\dfrac{(2r_2+1)\pi}{2n+1}$
$$\implies\prod_{r=0}^{n-1}\cos^2\dfrac{(2r+1)\pi}{2n+1}=\dfrac1{2^{2n}}$$
Now $\dfrac{(2r+1)\pi}{2n+1}>\dfrac\pi2\iff4r+2>2n+1\iff r>\dfrac{2n-1}4$
So, if $m=n-1-\left\lceil \dfrac{2n-1}4\right\rceil$
$$\implies(-1)^m\prod_{r=0}^{n-1}\cos\dfrac{(2r+1)\pi}{2n+1}=\dfrac1{2^n}$$
If $2r+1>\dfrac{2n+1}2, 0>\cos\dfrac{(2r+1)\pi}{2n+1}=-\cos\dfrac{2(n-r)\pi}{2n+1}$
$$\implies\prod_{m=1}^n\cos\dfrac{m\pi}{2n+1}=\dfrac1{2^n}$$
