Proving $n^2 \csc^2(nx) = \sum_{k=0}^{n-1}\csc^2\left(x+ k \frac{\pi}{n}\right)$ (without calculus?)

Question

I recently came across the following trigonometric identity in a test: $$ n^2 \csc^2(nx) = \sum_{k=0}^{n-1}\csc^2\left(x+ k \frac{\pi}{n}\right) $$ 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 $\csc^2 x$ 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 : $$ n \cot(nx) = \sum_{k=0}^{n-1} \cot\left(x+ k \frac{\pi}{n}\right) $$ 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)?

---

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

Let $n$ be an integer, then: $$\sin{nθ}=sinθ[\binom{n}{0}(2\cosθ)^{n-1}-\binom{n-1}{1}(2\cosθ)^{n-3}+\binom{n-2}{2}(2\cosθ)^{n-5}-...]$$ $$\cos{nθ}=\frac{1}{2}[(2cosθ)^{n}-\frac{n}{1}\binom{n-2}{0}(2\cosθ)^{n-2}+\frac{n}{2}\binom{n-3}{1}(2cosθ)^{n-4}-...]$$ You can get other identities by setting $$θ=\frac{π}{2}-ϕ; $$ and then consider different cases when n is even or odd and so on. Either way, in the second series $\cos{nθ}$ is in terms of powers of cosines, set $\cos{nθ}$ an arbitrary value, say $p$, then we also have that $\cos{(nϕ+2π)},\cos{(nϕ+4π)},...$ satisfy the equation, hence $\cos{(ϕ)},\cos{(ϕ+\frac{2π}{n})},\cos{(ϕ+\frac{4π}{n})},...$ are the roots of the equation on the right hand side, there are exactly n roots. Let$$ \cosθ=\frac{1}{q}$$ Upong making this substitution multiply by $q^n$ on both sides of the identity, then the roots of the new equation become $\sec{(ϕ)},\sec{(ϕ+\frac{2π}{n})},\sec{(ϕ+\frac{4π}{n})},...$, I'll consider the case when $n$ is odd, then $$\cos{nθ}=2^{n-1}(cosθ)^{n}-\frac{n}{1}\binom{n-2}{0}2^{n-3}(\cosθ)^{n-2}+...+(-1)^{\frac{n-1}{2}}n\cosθ$$ Making the said substitution and multiplying by $q^n$ yields: $$q^n\cos{nθ}=2^{n-1}-\frac{n}{1}\binom{n-2}{0}2^{n-3}q^2+...+(-1)^{\frac{n-1}{2}}nq^{n-1}$$It is a well known fact that the sum of roots is equal to the coefficient of the $q^{n-1}$ term divided by the coefficient of $q^{n}$, therefore: $$\sum_{k=1}^{n}\sec{(ϕ+\frac{(2k-2)π}{n})}=(-1)^{\frac{n-1}{2}}n\sec{nϕ}$$ Furthermore, $p_1^2+...+p_n^2=(p_1+...+p_n)^2-2\sum_{i≠j}p_ip_j$, but the sum of the roots taken two at a time is the coefficient of $q^{n-2}$ which is zero when $n$ is odd, thus when n is odd we have: $$\sum_{k=1}^{n}\sec^2{(ϕ+\frac{(2k-2)π}{n})}=n^2\sec^2{nϕ}$$ A similar derivation goes when n is even, for the sum of cosecants you may want to expand the sine in powers of sines and make the natural substitution $\sinθ=\frac{1}{q}$ and then use similar arguments with the roots so that the sum of its roots is a known coefficient, furthermore you can let $\sin^2{θ}=\frac{1}{q}$ and then let $p=q-1$ because $\cot^2{θ}+1=\csc^2{θ}$ so that by the same argument you can get the sum of cotangets and also of the cotangets squared, in the same manner one uses the cosines to build up the secant and then make use of the secant/tangent identity to find the sum of tangents. If you are interested in the the original series, you can derive them from the identity: $$\frac{\sin{θ}}{1-2xcos{θ}+x^2}=\sinθ+x\sin{2θ}+x^2\sin{3θ}+...ad inf$$ and: $$\frac{1-x^2}{1-2x\cos{θ}+x^2}=1+2x\cosθ+2x^2\cos{2θ}+... adinf$$ In both expressions you can expand the denominator as a geometric series if $x<1$ and the compare coefficients and finally get the expressions for $\sin{nθ}$ and $2\cos{nθ}$. (and by the way, the series in the beginning end whenever the binomial coefficient is either $\binom{n}{n-1}$ or $\binom{n}{n}$, and the signs alternate as plus, minus, plus,...) As you can see, the mathematical analysis way is easier and more convenient whilst using trigonometry can be tedious, the coefficients of this series behave nicely using the binomial notation, I hope it doesn't bother you posting this 2 years later but I figured it out I could employ trigonometry to somehow solve your problem.

Answer 2

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 $x=\frac{\pi}{2n}$.

Answer 3

$$\begin{align}\sin^2(n\theta)&=\left(\frac{e^{in\theta}-e^{-in\theta}}{2i}\right)^2\\&=-\frac14\left(-2+e^{i2n\theta}+e^{-i2n\theta}\right)\\&=\frac12\left(1-\sum_{k=0}^n\binom{2n}{2k}(-\sin^2\theta)^k(1-\sin^2\theta)^{n-k}\right)\\&=:P_n(\sin^2\theta). \end{align}$$ The $n$ numbers $s_k:=\sin^2\left(x+ k \frac\pi n\right)$ ($0\le k<n$) are therefore the roots of the polynomial $$P_n(s)-\sin^2(nx)=\sum_{k=0}^na_ks^k,$$ whence $$\sum_{k=0}^{n-1}\frac1{s_k}=-\frac{a_1}{a_0}=\frac{n^2}{\sin^2(nx)}.$$