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Calculus: $$T_1=\prod_{k=1}^{n-1} \cos\frac{k\pi}{2n}$$ and $$T_2=\prod_{k=1}^{n-1}\sin\frac{k\pi}{2n}$$

My tried:

I use Euler's formal: $$z_k=e^{i\frac{k\pi}{2n}}=\cos\frac{k\pi}{2n}+i\sin \frac{k\pi}{2n}$$

$$\to\left\{\begin{matrix}\cos\frac{k\pi}{2n}=\frac{1}{2}\left(z_k+\frac{1}{z_k}\right)\\\sin\frac{k\pi}{2n}=\frac{1}{2i}\left(z_k-\frac{1}{z_k}\right)\end{matrix}\right.$$

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  • $\begingroup$ Duplicate of this. $\endgroup$ – Lucian Jan 17 '14 at 18:48
  • $\begingroup$ You should see it, two completely different threads! $\endgroup$ – Iloveyou Jan 17 '14 at 19:08
  • $\begingroup$ @Lucian the linked question you give has denominator $n$ rather than $2n$ in the sine term : $sin\frac{k\pi}{n}$ in the linked question, but $\sin \frac{k\pi}{2n}$ in this question. (Don't know if it makes a big difference, but the question is not a duplicate directly.) $\endgroup$ – coffeemath Jan 17 '14 at 23:58
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First of all, $\displaystyle\cos\frac{k\pi}{2n}=\sin\left(\frac\pi2-\frac{k\pi} {2n}\right)=\sin\left(\frac{(n-k)\pi}{2n}\right)$

So, $T_1=T_2$

Now, $\displaystyle\sin\frac{k\pi}{2n}=\sin\left(\pi-\frac{k\pi}{2n}\right)=\cos\frac{(2n-k)\pi}{2n}$

So, $\displaystyle T_2=\prod_{k=1}^{n-1}\sin\frac{k\pi}{2n}=\prod_{k=n+1}^{2n-1}\cos\frac{k\pi}{2n}$

$\displaystyle T_2^2=\prod_{k=1,k\ne n}^{2n-1}\sin\frac{k\pi}{2n}=\prod_{k=1}^{2n-1}\sin\frac{k\pi}{2n}$ as $\sin\frac{n\pi}{2n}=1$

Observe that $0<\frac{k\pi}{2n}<\pi\implies\sin\frac{k\pi}{2n}>0$

$\displaystyle\implies T_2=+\sqrt{\prod_{k=1}^{2n-1}\sin\frac{k\pi}{2n}}$

Now from this,

$$\prod_{k=1}^{m-1}\sin\frac{k\pi}m=\frac m{2^{m-1}}$$

Here $m=2n$

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