# How prove this trigonometric identity

Show that

$$\sum_{k=0}^{n-1}\dfrac{\tanh{\left(x\dfrac{1}{n\sin^2{\left(\dfrac{2k+1}{4n}\pi\right)}}\right)}}{1+\dfrac{\tanh^2{x}}{\tan^2{\left(\dfrac{2k+1}{4n}\pi\right)}}}=\tanh{(2nx)}$$

Thank you ,This problem I take some hours,and at last I don't prove it

and This problem is from

This book have some same problem.all is not true? if not true,and we how find it or edit it somewhere?

Thank you achille hui,he told me this following maybe is true,Now How prove it?

$$\sum_{k=0}^{n-1}\dfrac{\dfrac{\tanh{x}}{n\sin^2{\left(\dfrac{2k+1}{4n}\pi\right)}}}{1+\dfrac{\tanh^2{x}}{\tan^2{\left(\dfrac{2k+1}{4n}\pi\right)}}}=\tanh{(2nx)}$$

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Are you sure you write down the expression correctly. It fails even when $n = 1$. For $n = 1$, the L.H.S simplifies to $\displaystyle \frac{\tanh(2x)}{1+\tanh(x)^2}$ which clearly differs from R.H.S. –  achille hui Sep 10 '13 at 12:39
Thank you,This problem is from this book:Table of Integrals,series,and Products,Seventh Edition page 39. –  china math Sep 10 '13 at 12:43
It is very likely the book copied it from somewhere else incorrectly. By the way, the correct expression seems to be $$\sum_{k=0}^{n-1} \frac{\frac{\tanh(x)}{n\sin^2(\frac{2k+1}{4n}\pi)} }{1 + \frac{\tanh^2 x}{\tan^2(\frac{2k+1}{4n}\pi)}} = \tanh(2nx)$$ At least for $n \le 4$, the corresponding graphs match. –  achille hui Sep 10 '13 at 12:51
oh,It's very well,Thank you –  china math Sep 10 '13 at 12:58
The $2^{nd}$ formula also suffers from the same problem. If you replace the $\tanh(\frac{x}{\cdots})$ by $\frac{\tanh x}{\cdots}$, then the graphs matches. I suspect the $3^{rd}$ and $4^{th}$ formula are having the same sort of mistakes. –  achille hui Sep 10 '13 at 13:05

$$\cosh(2nx) = T_{2n}(\cosh x)\quad\text{ and }\quad \cos(2nx) = T_{2n}(\cos x)$$
where $T_{2n}(z)$ is a Chebyshev polynomial of the first kind. Using the $2^{nd}$ relation above, it is clear the roots of $T_{2n}(x)$ has the form: $$\pm\cos(\frac{2k+1}{4n}\pi),\quad\text{ for } k = 0,\ldots, n-1$$
From this, we arrive following expansion of $\cosh(2n x)$:
$$\cosh(2n x ) = A \prod_{k=0}^{n-1}\left(\cosh^2 x - \cos^2(\frac{2k+1}{4n}\pi)\right)$$ for some constant $A$ we don't care. Taking logarithm, differentiate w.r.t $x$ and divide by $2n$ for both sides, we find: \begin{align} \tanh(2n x) = & \frac{1}{2n} \sum_{k=0}^{n-1}\frac{ 2\sinh x\cosh x}{\cosh^2 x - \cos^2(\frac{2k+1}{4n}\pi)}\\ = & \frac{1}{n}\sum_{k=0}^{n-1}\frac{ \sinh x\cosh x}{\sinh^2 x + \sin^2(\frac{2k+1}{4n}\pi)}\\ = & \frac{1}{n}\sum_{k=0}^{n-1}\frac{ \tanh x}{\tanh^2 x + \sin^2(\frac{2k+1}{4n}\pi)(1 - \tanh^2 x)}\\ \\ = & {\Large \sum_{k=0}^{n-1}\frac{\frac{\tanh x}{n \sin^2(\frac{2k+1}{4n}\pi)} }{1 + (\frac{1}{\sin^2(\frac{2k+1}{4n}\pi)} - 1 ) \tanh^2 x} }\\ \\ = & {\Large \sum_{k=0}^{n-1}\frac{\frac{\tanh x}{n \sin^2(\frac{2k+1}{4n}\pi)} }{1 + \frac{\tanh^2 x}{\tan^2(\frac{2k+1}{4n}\pi)}} }\\ \end{align}