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ganeshie8
 one year ago
show that \[\sum\limits_{k=1}^{n1}\csc^2\left(\dfrac{k\pi}{n}\right)=\dfrac{n^21}{3}\]
ganeshie8
 one year ago
show that \[\sum\limits_{k=1}^{n1}\csc^2\left(\dfrac{k\pi}{n}\right)=\dfrac{n^21}{3}\]

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thomas5267
 one year ago
Best ResponseYou've already chosen the best response.2Any identity relating to products of sine? We can rearrange the sum as follows: \[ \sum_{k=1}^{n1}\csc^2\left(\dfrac{k\pi}{n}\right)=\frac{\sum_{i=1}^{n1}\prod_{k\neq i}^{n1}\sin^2\left(\frac{k\pi}{n}\right)}{\prod_{k=1}^{n1}\sin^2\left(\frac{k\pi}{n}\right)} \]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.2That is, the numerator is the coefficient of the \(x\) of the polynomial \(p(x)=\prod\limits_{i=1}^{n1}\left(x+\sin^2\left(\dfrac{k\pi}{n}\right)\right)\) and the denominator is the constant term of \(p(x)\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{k=1}^{n1} \csc^2(\frac{k\pi}{n})=\sum_{k=1}^{n1}1+\sum_{k=1}^{n1} \cot^2(\frac{k\pi}{n})=(n1)+\sum_{k=1}^{n1} \cot^2(\frac{k\pi}{n})\] \[(n1)+\sum_{k=1}^{n1}(\frac{1+\cos(\frac{2k\pi}{n})}{1\cos(\frac{2k\pi}{n})})\] No idea, just messing around a bit :p

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.2Two relevant links on Math Stack Exchange. They suggested using Chebyshev polynomial of the first and second kind to solve the problem. I cannot make this work. Maybe someone can on here. https://math.stackexchange.com/questions/122836/sumofthereciprocalofsinesquared https://math.stackexchange.com/questions/45144/provingfrac1sin2fracpi14frac1sin2frac3pi14/
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