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$$ A_{n} = \prod_{k=1}^{n} \cos\left(\frac{k\pi}{2n+1}\right) $$ Let $x_k = \cos\left(\frac{k\pi}{2n+1}\right)$. Let $T_{2n+1}$ the $2n+1$st Chebyshev polynomial of the first kind, which satisfies: $$\begin{cases} T_{2n+1}(\cos(\theta)) = \cos((2n+1)\theta) \\ T_{2n+1}(-x) = -T_{2n+1}(x) \\ \mathrm{deg}(T_{2n+1}) = 2n+1 \\ \textrm{leading coefficient}=2^{2n} \end{cases}$$ We have: $$T_{2n+1}(\pm x_k) = \pm(-1)^k\textrm{ where }k\in[\![0; 2n]\!]$$ Thus $P = T_{2n+1}^2 - 1$ has $4n+2$ zeroes which are the $\pm x_k$ for $0 \le k \le 2n$, and a leading coefficient of $2^{4n}$. We also have $x_{2n+1-k} = -x_k$, thus the product of roots of $P$ is: $$\begin{align*} \frac{P(0)}{2^{4n}} &= \prod_{k=0}^{2n} x_k\cdot \prod_{k=0}^{2n}(-x_k) \\ &= -\left(\prod_{k=0}^{2n} x_k\right)^2 \\ &= -\left(-\prod_{k=1}^{n} x_k\right)^4 \\ \implies A_{n} &= \frac{\sqrt[4]{1-T_{2n+1}(0)^2}}{2^n} \\ &= \boxed{\frac{1}{2^{n}}} \end{align*}$$
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Thu, 05 Sep 2024 15:19 GMT