TITLE: Evaluation of $\cos\left(\frac{\pi}{2n}\right)\cdot \cos\left(\frac{2\pi}{2n} \right)....\cos\left(\frac{(n-1)\pi}{2n}\right)$
QUESTION [5 upvotes]: Evaluation of $$\lim_{n\rightarrow \infty}\left(\tan \frac{\pi}{2n}\cdot \tan \frac{2\pi}{2n}\cdot \tan \frac{\pi}{3n}\cdot ...............\tan \frac{(n-1)\pi}{2n}\right)^{\frac{1}{n}} = $$ without using Limit as a sum.

$\bf{My\; Try::}$ Using the formula $$\displaystyle \sin\left(\frac{\pi}{n}\right)\cdot \sin\left(\frac{2\pi}{n} \right)....\sin\left(\frac{(n-1)\pi}{n}\right) = \frac{n}{2^{n-1}}$$
Replace $n\rightarrow 2n$
$$\displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(2n-1)\pi}{2n}\right) = \frac{2n}{2^{2n-1}}$$
Now How can I calculate $$\displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(n-1)\pi}{2n}\right)$$
and also How can I calculate $$\displaystyle \cos\left(\frac{\pi}{2n}\right)\cdot \cos\left(\frac{2\pi}{2n} \right)....\cos\left(\frac{(n-1)\pi}{2n}\right)$$
Help required, Thanks

REPLY [2 votes]: First we have
\begin{align}
\lim_{n\to \infty}\frac1{n}\ln{\left(\tan\frac{\pi}{2n}.\tan\frac{2\pi}{2n}\cdots\tan\frac{(n-1)\pi}{2n}\right)}
&=\lim_{n\to \infty}\frac1{n}\sum\limits_{k=0}^{n-1}\ln{\left(\tan\frac{k\pi}{2n}\right)}
\\
&=\frac{2}{\pi}\int_0^{\pi/2}\ln{\tan{x}}\:dx
\\
&=\frac{2}{\pi}\left(\int_0^{\pi/2}\ln\sin(x)dx -\int_0^{\pi/2}\ln\cos(x)dx\right)
\\
&=0
\end{align} 
The last step is through $y=\pi/2-x$.
Hence
$$
\lim_{n\to\infty}\left(\tan\frac{\pi}{2n}.\tan\frac{2\pi}{2n}\cdots\tan\frac{(n-1)\pi}{2n}\right)^{\frac1{n}}=e^0=1
$$