Question: Compute the value of the product $$\prod^{100}_{n=1} \left(5-4 \cos \left(\frac{\pi (2n-1)}{100} \right) \right)$$
I began by considering $$a_k=\prod^{k}_{n=1} \left(5-4 \cos \left(\frac{\pi (2n-1)}{k} \right) \right)$$ for smaller values of $k$.
- $a_1=3$
- $a_2=25$
- $a_3=81$
- $a_4=289$
- $a_5=1089$
This led me to conjecture that $a_k=(2^k+1)^2$. However, I am unsure how to go about proving this.
I first began by rewriting $$a_k=4^k \cdot \prod^{k}_{n=1} \left(\frac{5}{4}-\cos \left(\frac{\pi (2n-1)}{k} \right) \right)$$
I then thought of interpreting this as $P(5/4)$, where $$P(x)=\prod^{k}_{n=1} \left(x-\cos \left(\frac{\pi (2n-1)}{k} \right) \right)$$
Initially, I thought this looked similar to De Moivre's theorem, since $P(x)$ is equal to the product of $\Re \left(x-e^{\frac{i\pi(2n-1)}{k}} \right)$. However, when multiplying imaginary terms by each other, the result is real, and they don't seem to cancel nicely.
Best Answer
Denote $\xi_k=e^{\frac{2\pi i}k}$. Note that \begin{align*} x^k+1&=x^k-\left(e^{-\frac{i\pi}k}\right)^k=\prod_{n=1}^k\left(x-e^{-\frac{i\pi}k}\xi_k^n\right)=\prod_{n=1}^k\left(x-e^{i\frac{2n-1}{k}\ \ \pi}\right),\qquad x\in\mathbb R. \end{align*} Multiplying the above expression with its conjugate gives that \begin{align*} \left(x^k+1\right)^2&=\prod_{n=1}^k\left(x-e^{i\frac{2n-1}{k}\ \ \pi}\right)\prod_{n=1}^k\left(x-e^{-i\frac{2n-1}{k}\ \ \pi}\right)\\ &=\prod_{n=1}^k\left(x^2+1-2x\cos\left(\frac{2n-1}k\pi\right)\right),\qquad x\in\mathbb R. \end{align*} Taking $x=2$ we obtain $$\prod_{n=1}^k\left(5-4\cos\left(\frac{2n-1}k\pi\right)\right)=\left(2^k+1\right)^2.$$