# Handle wanted for a really neat trig identity [closed]

Is there a handle for the trig identity used in giving a “fairly direct and unsurprising” derivation of Euler’s determination of zeta of 2?

In an article on the web (whose title I think is misleading, and so I’m not going to give it publicity), dated December 6, 2002, Dan Kalman, of American University, proves this identity, and uses for the aforesaid purpose, but does not give it a handle (i.e., name) for easy reference. Can anyone supply it, or, if none yet exists, suggest one?

The identity in question is: For $\omega=\pi/(2m+1)$, $$\cot^2(\omega)+\cot^2(2\omega)+\cot^2(3\omega)+\cdots+\cot^2(m\omega) = \frac{m(2m-1)}{3}.$$

-

## closed as not a real question by Gerry Myerson, Henning Makholm, Benjamin Lim, t.b., Asaf KaragilaMar 24 '12 at 20:09

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

The article in question is this. Just because you feel a title is misleading is no justification for referencing it without giving any sort of link or self-contained summary/snippet. –  anon Mar 10 '12 at 5:09
This is just silly. Why not just provide the identity here? You are just making harder for people to give an answer by not giving either the identity, or a reference to the article, and forcing people to do some extra work just to understand what in the world you are talking about. –  Aryabhata Mar 10 '12 at 5:12
@anon: If I didn't give a link, then how did you find it? –  Hexagon Tiling Mar 10 '12 at 5:14
@HexagonTiling: If you cannot do the formatting, you can atleast refer to the identity by giving a link to the article, mentioned the page number and telling us how we can get to it. Someone will edit it in for you. If you cannot be bothered to do it, it is very rude(IMO) to expect others to do that for you. People give their time and knowledge here for free and here you are, taking them for granted. Sorry if that sounds harsh. –  Aryabhata Mar 10 '12 at 5:28
HexagonTiling, you are being very rude by saying what you said in your last comment. My suggestion is: do not ask questions if you are not motivated enough to put the (very little) time and effort needed to do it minimally properly. Your "maybe then I'll do my homework" is simply disrespectful in the context of this site. –  Mariano Suárez-Alvarez Mar 10 '12 at 6:08