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Evaluate using complex numbers: $$\prod^{n}_{k=1}\cos\left(\frac{k\pi}{m}\right)$$ where $m=2n+1$.

$\bf{My\; Try::}$ Let $\displaystyle P = \prod^{n}_{k=1}\cos\left(\frac{k\pi}{m}\right).$ Now let $\displaystyle \cos \left(\frac{k\pi}{m}\right) = \frac{e^{\frac{ik}{m}}+e^{\frac{-ik}{m}}}{2}$ and Let $e^{\frac{i\pi}{m}} = \omega$

So we get $$P = \frac{1}{2^n}\left\{(\omega+\omega^{-1})(\omega^2+\omega^{-2})\cdots(\omega^n+\omega^{-n})\right\}$$

So $$P = \frac{(1+\omega^2)(1+\omega^4)\cdots(1+\omega^{2n})}{2^n\cdot \omega^{\frac{n(n+1)}{2}}}$$

Now How can i solve after that, Help required, Thanks

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Note that $P$ is a positive real number. Hence, $$P^2=|P|^2=P\,\bar{P}= \left(\frac{(1+\omega^2)(1+\omega^4)\cdots(1+\omega^{2n})}{2^n \omega^{\frac{n(n+1)}{2}}}\right)\left(\frac{(1+\bar\omega^2)(1+\bar\omega^4)\cdots(1+\bar\omega^{2n})}{2^n \bar\omega^{\frac{n(n+1)}{2}}}\right)\,.$$ As $\omega\bar\omega=|\omega|^2=1$ and $\bar\omega^{2j}=\omega^{2n+1-2j}$ for every $j=1,2,\ldots,n$, we have $$P^2=\frac{\prod_{k=1}^{2n}\,\left(1+\omega^k\right)}{2^{2n}}\,.$$ Now, the polynomial $f(z):=z^{2n}+z^{2n-1}+\ldots+z+1$ can be factorized as $$f(z)=\prod_{k=1}^{2n}\,\left(z-\omega^k\right)\,.$$ That is, $$P^2=\frac{(-1)^{2n}\,f(-1)}{2^{2n}}=\frac{1}{2^{2n}}\,.$$ This shows that $P=\dfrac{1}{2^n}$.

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By using Chebyshev polynomials of the second kind, from $$ \frac{U_n(x)}{2^n} = \prod_{k=1}^{n}\left(x-\cos\frac{k\pi}{n+1}\right)\tag{1}$$ it follows that the wanted product just depends on $\frac{U_{2m}(0)}{2^{2m}}$, where $U_{2m}(0)=(-1)^m$ follows from: $$ U_{n+2}(t) = 2t\cdot U_{n+1}(t)-U_n(t).\tag{2} $$

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