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Is this a trigonometric identity? $$ \sum_{k=1}^n \frac 2 {1 - \cos\left( \frac{2k-1} n \cdot\pi \right)} = n^2 $$

If I'm not mistaken, I can prove this when $n$ is a power of $2$, by induction on the exponent. Numerical evidence suggests it is true of positive integers $n$ in general. Can it be proved?

Blue
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Michael Hardy
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    This may or may not help: Set $$f(x)=\prod_{k=1}^{n}\left(x-\cos\tfrac{(2k-1)\pi}{n}\right),$$ so that $$\frac{f'(x)}{f(x)}=\sum_{k=1}^{n}\frac1{x-\cos\tfrac{(2k-1)\pi}{n}},$$ and your sum is given by $S=2f'(1)/f(1)$. – clathratus Jan 21 '20 at 06:50
  • Same as here, we have $$\sum_{k = 1}^n \frac 2 {1 - \cos \frac {(2 k - 1) \pi} n} = \sum_{k = 1}^n \operatorname* {Res}{z = e^{(2 k - 1) \pi i /n}} \frac {4 n} {(z - 1)^2 (z^n + 1)} = -\operatorname* {Res}{z = 1} \frac {4 n} {(z - 1)^2 (z^n + 1)}.$$ – Maxim Jan 23 '20 at 01:23

1 Answers1

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We need $$\sum_{k=1}^n\csc^2\dfrac{(2k-1)\pi}{2n}=n+\sum_{k=1}^n\cot^2\dfrac{(2k-1)\pi}{2n}$$

Now using Trig sum: $\tan ^21^\circ+\tan ^22^\circ+\cdots+\tan^2 89^\circ = \text{?}$

$$\cot2nx=\dfrac{\binom{2n}0c^{2n}-\binom{2n}2c^{2n-2}+\cdots}{\binom{2n}1c^{2n-1}+\cdots}$$ where $c=\cot x$

Now if $\cot2nx=0,2nx=(2m+1)\dfrac\pi2$ where $m$ is any integer

$$x=\dfrac{(2m+1)\pi}{2n}$$ where $m=0,1,2\cdots,2n-1$

SO, the roots of $$\binom{2n}0c^{2n}-\binom{2n}2c^{2n-2}+\cdots=0$$ are $\cot\dfrac{(2m+1)\pi}{2n};m=0,1,2\cdots,2n-1$

So, the roots of $$\binom{2n}0d^n-\binom{2n}2d^{n-1}+\cdots=0$$ are $\cot^2\dfrac{(2m+1)\pi}{2n};m=0,1,2\cdots,n-1$

$$\implies\sum_{k=1}^n\cot^2\dfrac{(2k-1)\pi}{2n}=\dfrac{\binom{2n}2}{\binom{2n}0}$$

Have I missed soemthing?

Michael Hardy
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lab bhattacharjee
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