## Summations in $\tan^2$

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Hey all,

I was just wondering if anyone had come across the following identities, valid for $m\in\mathbb{N}$. I've used Abramowitz and Stegun, Maple, Mathematica etc but can't find them anywhere. I can prove these, though they happen 'accidentally' from a method which I am already looking at. Anyway the identities are

$$\sum_{k=1}^m \left(\tan\left(\frac{\pi(2k-1)}{4m}\right)\right)^2=m(2m-1) \hspace{4mm} \textrm{and} \hspace{4mm} \sum_{k=1}^m \left(\tan\left(\frac{\pi k}{2m+1}\right)\right)^2=m(2m+1)$$

and

$$\sum_{k=1}^m \left(\tan\left(\frac{\pi(2k-1)}{4m}\right)\right)^4=\frac{1}{3}m(2m-1)(4m^2+2m-3) \hspace{4mm} etc$$

There are other identities for all even powers but I haven't worked them out yet as I thought that there might not be any point if there are known results for these summations. It would be cool if there were lists of such identities, or even a general formula, as this would provide me with many useful references indeed!

Many thanks on Christmas!

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How do you derive these identities? – Anixx Dec 25 2010 at 15:19
seems you confused m and n in the last identity – Anixx Dec 25 2010 at 15:20
Not much of a simplification, but your first sum can be "cut in half": for even $m$, your sum is the same as $$\sum_{k=1}^{\lfloor m/2\rfloor}\left(\tan^2\left(\frac{\pi}{4m}(2k-1)\right)+\cot^2\left(\frac{\pi}{4m}(2k-1)\right)\right)$$ ; for odd $m$, add 1. – J. M. Dec 25 2010 at 15:22
(and something similar can be done for the other sums) – J. M. Dec 25 2010 at 15:24

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You'll find this and a list of further identities here emis.de/journals/HOA/IJMMS/30/3185.pdf – dke Dec 25 2010 at 15:48
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