How to find $$\cot^2\frac{\pi}{2m+1}+\cot^2\frac{2\pi}{2m+1}+\cot^2\frac{3\pi}{2m+1}+\ldots+\cot^2\frac{m\pi}{2m+1}?$$ Of course, the number $m$ is assumed to be a positive integer.
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I'd rewrite $\cot^2 x = \csc^2 x - 1$. – Thomas Andrews Feb 20 '15 at 19:39
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3@ Thomas Andrews : So then? – Markiyan Hirnyk Feb 20 '15 at 19:44
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1Do you know $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$, $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$? If so, you use it there, expand what you get as series of powers of $e^{ix}$ and sum the geometric progressions you get. In the process you will need to decompose $\frac{X}{X^2-2X-1}$ in simple fractions and the expansion $\frac{1}{1-X}=\sum_k X^k$. – Tom Feb 20 '15 at 19:47
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See this link http://math.stackexchange.com/questions/265229/prove-that-cot2-pi-7-cot22-pi-7-cot23-pi-7-5. – xpaul Feb 20 '15 at 19:49
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@ xpaul : That is not it, only a related question. – Markiyan Hirnyk Feb 20 '15 at 19:52
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@ Tom : Can you kindly realize your idea as an answer? – Markiyan Hirnyk Feb 20 '15 at 20:01
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@MarkiyanHirnyk, use the same idea for general $m$ (in the link $m=3$). – xpaul Feb 20 '15 at 20:25
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@MarkiyanHirnyk, See http://math.stackexchange.com/questions/1141829/summation-of-a-trigonometric-series – lab bhattacharjee Feb 21 '15 at 05:04
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Here is a rather elementary and simple way to sum this series:
We know from De Moivre's Formula that
$$(\cos\theta+\iota\sin\theta)^n=\cos n\theta+\iota\sin n\theta$$
Expanding the LHS using Binomial Theorem,
$$\binom{n}{0}\cos^n\theta+\binom{n}{1}\cos^{n-1}\theta(\iota\sin\theta)+\cdots+\binom{n}{n}\iota^n\sin^n\theta=\cos n\theta+\iota\sin n\theta$$ Equating imaginary parts,
$$\binom{n}{1}\cos^{n-1}\theta(\sin\theta)-\binom{n}{3}\cos^{n-3}\theta(\sin\theta)^3+\binom{n}{5}\cos^{n-5}\theta(\sin\theta)^5-\cdots=\sin n\theta$$ Now let $n=2m+1$,then:
$$\binom{2m+1}{1}\cos^{2m}\theta(\sin\theta)-\binom{2m+1}{3}\cos^{2m-2}\theta(\sin\theta)^3+\cdots=\sin (2m+1)\theta$$ Dividing by $\sin^{2m+1}\theta$, we have a result:
$$\binom{2m+1}{1}(\cot^2\theta)^m-\binom{2m+1}{3}(\cot^2\theta)^{m-1}+\cdots={\sin (2m+1)\theta \over \sin^{2m+1}\theta}\tag{i}$$
Now consider the following equation of the $m^{th}$ degree:
$$\binom{2m+1}{1}x^m-\binom{2m+1}{3}x^{m-1}+\binom{2m+1}{5}x^{m-2}-\cdots=0$$
Making use of the preceding result (i), we can say that $\cot^2{\pi \over 2m+1}$,$\cot^2{2\pi \over 2m+1},\ldots,\cot^2{m\pi \over 2m+1}$ are the $m$ roots of this equation.
Using Vieta's formulas, the sum of roots of this equation (which is the required sum) must be this: $${2m+1 \choose 3}\over{2m+1 \choose 1}$$ Which on simplification reduces to this: $${2m^2-m}\over{3}$$
najayaz
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