<s>[INST] Evaluation of $\cos\left(\frac{\pi}{2n}\right)\cdot \cos\left(\frac{2\pi}{2n} \right)....\cos\left(\frac{(n-1)\pi}{2n}\right)$ 
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
 [/INST] For fixed $n$, call the expression $A$, and take its $\log[.] $ to get:
$$ \log (A)=\frac{1}{n} \left(\log\tan \frac{\pi}{2n}+ \tan \frac{2\pi}{2n}+ .............. + \tan \frac{(n-1)\pi}{2n}\right) $$
$$ =\frac{1}{n} \sum_{k=1}^{n-1}\log\tan (\frac{\pi k}{2n})$$
Doesn't it look familiar?! Yes, Riemann sum as in the definition of definite integrals. Here, if $\ k/n$ is seen as our discretized parameter $x$, then
$$\lim_{n\rightarrow \infty} \log (A) = \int_0^1\log\tan(\frac{\pi }{2}x)dx$$
Once this value is calculated, $A$ will be the exponential of it. But how do we get $$ \int_0^1\log\tan(x)dx \ ?$$
Despite its nasty look, there seems to be hope to find an antiderivative. But I don't know right away.
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