Trigonometric summation involving the squares of cosecant terms.

Question

I recently came across the following trigonometric identity in a test: $$ \bigl[n \csc(nx)\bigr]^2 = \sum_{k=0}^{n-1} \bigl[\csc(x+ k \pi/n) \bigr]^2 $$ The question was to prove the result for any natural number $n$.

What would be a good way to approach this problem?

My initial line of thought was that since $(\csc x)^2$ is the negative of the derivative of $\cot x$, the sum of the above series could be evaluated by differentiating a series involving the cotangent. Hence the question is equivalent to showing that : $$ n \cot(nx) = \sum_{k=0}^{n-1} \cot(x+ k \pi/n) $$ Taking the derivative of both sides with respect to the variable $x$ and multiplying the resulting equation by $-1$, we arrive at the required result. Although this does look simpler, I couldn't find a way to calculate the new sum. Could logarithms be used for this?

Does this method work on further simplification? Or is there an alternative route to the answer (involving, for instance, complex numbers)?

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EDIT:

It turns out that the method does indeed work, as explained in this answer, where the second summation has been calculated using basic trigonometric expansions and a bit of algebra. Nevertheless, is there a different way to prove the identity without using calculus? Or even better(ideally), from trigonometry alone? Invoking calculus in a trig problem of this sort seems a tad unnatural, unintuitive and unappealing to me.

Answer 1

The first trigonometric sum, related with $\csc^2$, was used by Cauchy to provide an elementary proof of the identity $\zeta(2)=\sum_{n\geq 1}\frac{1}{n^2}=\frac{\pi^2}{6}$. Wikipedia gives his derivation in full detail: this answers your question about a Calculus-free proof.

Under a modern perspective (in which Complex Analysis is not an enemy but just a tool), such sums can be computed through Herglotz' trick since $\cot$ and $\csc^2$ are associated with simple Eisenstein series. A good starting point is Weierstrass' product for the sine function: $$ \frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right)\tag{1} $$ If we apply $\frac{d}{dz}\log(\cdot)$ to both sides of $(1)$, we get: $$ -\frac{1}{z}+\pi\cot(\pi z)=\sum_{n\geq 1}\frac{2z}{z^2-n^2},\qquad \cot(\pi z)=\frac{1}{\pi}\sum_{m\in\mathbb{Z}}'\frac{1}{z-m} \tag{2}$$ where the prime mark stands for: the series has to be considered in a symmetric sense. From $(2)$ we have that $\cot(z)$ is a meromorphic function over the complex plane, with simple poles with residue $1$ at every integer. By differentiating again: $$ \csc^2(\pi z)=\frac{1}{\pi^2}\sum_{m\in\mathbb{Z}}'\frac{1}{(z-m)^2}\tag{3} $$ we get that $\csc^2(z)$ is a meromorphic function with double poles at every integer, with every double pole behaving in the same way. Your identities can now be proved by analyzing the poles of the involved LHS/RHS, checking that the behaviour at the singular points is the same, then proving LHS$=$RHS for just a specific value of $x$, like $x=\frac{\pi}{2n}$.