Finding $$\mu=\prod_{k=1}^{n-1}\cos\frac{2k\pi}n$$ I thought $$z^n=1=e^{i2\pi}\implies z=\cos\frac{2k\pi}n+i\sin\frac{2k\pi}n\quad k\in\{1,2,...,n-1\}$$ Now we have: $$\begin{align}\mu&=\prod_{k=1}^{n-1}\frac{z+\bar z}{2}\\&=\prod_{k=1}^{n-1}\frac{e^{i\theta}+ e^{-i\theta}}{2}\\&=\frac1{2^{n-1}}\prod_{k=1}^{n-1}e^{-i\theta}\prod_{k=1}^{n-1}(1+ e^{2i\theta})\\ \\&=\frac1{2^{n-1}}e^{-i\frac{2\pi}n(1+2+...(n-1))}\prod_{k=1}^{n-1}(1+ e^{2i\theta})\\ \\&=\frac1{2^{n-1}}e^{-i(n-1)\pi}\prod_{k=1}^{n-1}(1+ e^{2i\theta})\\ \\&=\frac1{2^{n-1}}(-1)^{n-1}\prod_{k=1}^{n-1}(1+ e^{2i\theta})\end{align}$$ Now since $(e^{2i\theta})^{n/2}=1$: $$\Theta(z)=z^{n/2}-1=\prod_{k=1}^{n-1}(z-e^{2i\theta})$$ At $(-1)$: $$\Theta(-1)=(-1)^{n-1}\prod_{k=1}^{n-1}(1+e^{2i\theta})=(-1)^{n/2}-1=e^{in\pi/2}-1=i\sin n\pi/2$$ It turns out something imaginary (but it's not clearly) also I would like to work on: $$\nu=\prod_{k=1}^{n-1}\sin\frac{2k\pi}n=\prod_{k=1}^{n-1}\frac{e^{i\theta}- e^{-i\theta}}{2i}$$ Similarly: $$\nu=\frac1{(2i)^{n-1}}(-1)^{n-1}\prod_{k=1}^n(e^{2i\theta}-1)=\frac1{(2i)^{n-1}}(-1)^{n-1}(-1)^{n-1}\Theta(1)=\frac1{2^{n-1}i\sin (n-1)\pi/2}0=0$$ But this is also not zero? The earlier $\eta$ is obviously: $$\prod_{k=1}^{n-1}\tan\frac{2k\pi}n=\eta=\frac{\mu}{\nu}$$
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This is a related question. – mathlove May 12 '15 at 13:27
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@mathlove can you help with my mistake? – RE60K May 12 '15 at 13:29
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@ADG Can you please explain why $z^{n/2}-1=\prod\limits_{k=1}^{n-1}(z-e^{2i\theta})$ ? – Alexey Burdin May 12 '15 at 13:39
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@AlexeyBurdin as $(e^{i\theta})^n=1$ so $(e^{2i\theta})^{n/2}=1$ therefore $e^{2i\theta}$ are the roots of $z^{n/2}=1$ – RE60K May 12 '15 at 13:41
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@ADG It turns out not. There is degree $n/2$ "polynomial" on the left and degree $n-1$ polynomial (respect to $z$) on the right. – Alexey Burdin May 12 '15 at 13:51
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@AlexeyBurdin oh that's the mistake, then can you help improve it? – RE60K May 12 '15 at 13:52
1 Answers
$$ \begin{align} 2^{n-1}\prod_{k=1}^{n-1}\cos\left(\frac{2k\pi}n\right) &=\prod_{k=1}^{n-1}\left(1+e^{4\pi ik/n}\right)e^{-2\pi ik/n}\\ &=(-1)^{n-1}\left[\prod_{k=1}^{n-1}\left(-z^2+e^{4\pi ik/n}\right)\right]_{\large z=i}\\ &=\left[\prod_{k=1}^{n-1}\left(z^2-e^{4\pi ik/n}\right)\right]_{\large z=i}\\ &=\left[\prod_{k=1}^{n-1}\left(z-e^{2\pi ik/n}\right)\prod_{k=1}^{n-1}\left(z+e^{2\pi ik/n}\right)\right]_{\large z=i}\\[3pt] &=\left[\frac{z^n-1}{z-1}\frac{z^n-(-1)^n}{z+1}\right]_{\large z=i}\\[6pt] &=\frac{(i^n-1)(i^n-(-1)^n)}{-2}\\[9pt] &=\frac{2(-1)^n-i^n(1+(-1)^n)}{-2}\\[12pt] &=\cos(n\pi/2)-(-1)^n \end{align} $$ Therefore, $$ \begin{align} \prod_{k=1}^{n-1}\cos\left(\frac{2k\pi}n\right) &=\frac{\cos(n\pi/2)-(-1)^n}{2^{n-1}}\\[3pt] &=\frac{\cos(n\pi/2)-\cos(n\pi)}{2^{n-1}}\\[6pt] &=\frac{\sin(n\pi/4)\sin(3n\pi/4)}{2^{n-2}} \end{align} $$
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1I always get notifications starting with "$$\begin{align}... "and I intutively know it's you robjohn, thanks!. – RE60K May 12 '15 at 15:18
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see http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Identities_without_variables, and cos(npi/2)=0 too! – RE60K May 12 '15 at 15:22
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no you're not, that's me just relating it. and ok only odd multiples go to zero sorry – RE60K May 12 '15 at 15:29
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I found this: $(-1)^{-n}2^{1-n}\left(\cos\left(\frac{n\pi}{2}\right)-1\right)$ – Jan Eerland May 12 '15 at 16:22
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@JanEerland: that differs by the sign of $\cos\left(\frac{n\pi}2\right)$ from my answer. Have you checked that numerically? – robjohn May 12 '15 at 16:53
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@JanEerland: Oh, the sign only differs when $n$ is odd, and then $\cos\left(\frac{n\pi}2\right)=0$. So our answers actually do agree. – robjohn May 12 '15 at 17:13