# A conjectural trigonometric identity

Recently, I formulated the following conjecture which seems novel.

Conjecture. For any positive odd integer $$n$$, we have the identity $$\sum_{j,k=0}^{n-1}\frac1{\cos 2\pi j/n+\cos 2\pi k/n}=\frac{n^2}2.\tag{1}$$

Using Galois theory, I see that the sum is a rational number. The identity $$(1)$$ has some equivalent versions, for example, $$\sum_{0\le j and $$\sum_{1\le j It is easy to check $$(1)$$-$$(3)$$ numerically.

Question. How to prove the conjecture?

• If we put $n=8$ in the sum under consideration, then up to Maple and Mathematica we face with division by zero. E.g. restart; n := 8; add(add(1/(cos(2*Pi* j/n) + cos(2*Pi* k/n)), k = 0 .. n - 1), j = 0 .. n - 1); Error, numeric exception: division by zero – user64494 Aug 1 at 19:20
• @user64494 but the conjecture was about odd integers – Fedor Petrov Aug 1 at 20:59
• @Fedor Petrov: Thank you. Where do the below proofs use it? – user64494 Aug 2 at 6:43
• @user64494 everywhere:) for even $n$ we indeed have zero denominators (say, for $j=0,k=n/2$ and others), it is senseless. In both below solutions the logarithmic derivative of some polynomial at some point is calculated, and for even $n$ this point would be a root of the polynomial. – Fedor Petrov Aug 2 at 8:28
• The identity (1) is a special case of a much more general result, see Theorem 1.2 of my preprint arxiv.org/abs/1908.02155. – Zhi-Wei Sun Aug 8 at 3:11

Actually this sum quickly factorizes: using $$\cos (x+y)+\cos (x-y)=2\cos x\cos y$$ and denoting $$(j+k)\cdot \frac{n+1}2=a,(j-k)\cdot \frac{n+1}2=b$$ we get $$\cos 2\pi j/n+\cos 2\pi k/n=2\cos 2\pi a/n \cdot \cos 2\pi b/n.$$ Here $$j, k, a, b$$ may be considered as residues modulo $$n$$ ($$\cos 2\pi a/n$$ depends only on $$a$$ modulo $$n$$), and the map $$(j,k)\mapsto (a,b)$$ is bijective on $$(\mathbb{Z}/n\mathbb{Z})^2$$, thus $$2\sum_{j,k=0}^{n-1}\frac1{\cos 2\pi j/n+\cos 2\pi k/n}=\sum_{a,b=0}^{n-1}\frac1{\cos 2\pi a/n\cdot \cos 2\pi b/n}=\left(\sum_{a=0}^{n-1}\frac1{\cos 2\pi a/n}\right)^2.$$ It remains to calculate the sum $$\sum_{a=0}^{n-1}\frac1{\cos 2\pi a/n}$$. Denoting $$\omega=e^{2\pi i/n}$$ we get $$\sum_{a=0}^{n-1}\frac1{\cos 2\pi a/n}=\sum_{a=0}^{n-1}\frac{2\omega^a}{1+\omega^{2a}}= \sum_{a=0}^{n-1}\frac{\omega^a+\omega^{(2n+1)a}}{1+\omega^{2a}}=\\ \sum_{a=0}^{n-1} (\omega^a-\omega^{3a}+\dots+(-1)^{(n-1)/2}\omega^{na}+\dots+\omega^{(2n-1)a})=(-1)^{(n-1)/2}n,$$ since the geometric progressions $$\sum_{a=0}^{n-1} \omega^{ma}$$ sums up to $$\sum_{a=0}^{n-1} \omega^{ma}=\begin{cases}0& \text{for}\, m\, \text{not divisible by}\, n\\ n& \text{for}\, m\, \text{divisible by}\, n. \end{cases}$$
Let $$T_n$$ be the $$n$$-th Chebyshev polynomial, so that $$T_n(x)-1=\prod_{j=1}^n(x-\cos 2\pi j/n).$$ Taking a logarithmic derivative we have $$\sum_{j=0}^{n-1}\frac{1}{x-\cos 2\pi j/n}=\frac{T_n'(x)}{T_n(x)-1}.$$ For $$x=-\cos 2\pi k/n$$, we easily find $$T_n(x)=-1$$. We therefore have $$\sum_{j=0}^{n-1}\frac{1}{\cos 2\pi k/n+\cos 2\pi j/n}=\frac{T_n'(-\cos 2\pi k/n)}{2}.$$ Now the derivative of $$T_n$$ is equal to $$nU_{n-1}$$, where $$U_{n-1}$$ is the Chebyshev polynomial of the second kind, satisfying $$U_{n-1}(\cos t)=\frac{\sin nt}{\sin t}$$.
For $$k\neq 0$$, at $$t=2\pi k/n+\pi$$, so that $$\cos t=-\cos 2\pi k/n$$, we get $$U_{n-1}(-\cos 2\pi k/n)=\frac{\sin(2\pi k+n\pi)}{\sin(2\pi k/n+\pi)}=0$$. For $$k=0$$, we have to take a limit at $$t\to\pi$$ and we find $$U_{n-1}(1)=n$$. Therefore we get $$\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}\frac{1}{\cos 2\pi k/n+\cos 2\pi j/n}=\frac{T_n'(1)}{2}=\frac{n^2}{2}.$$
Denote $$\omega=e^{2\pi i/n}$$, then $$\Omega=\{\omega^k:k=0,\ldots,n-1\}$$ is the set of roots of the polynomial $$f(t):=t^n-1$$, $$-\Omega=\{-\omega^k:k=0,\ldots,n-1\}$$ the set of roots of $$g(t):=t^n+1$$. We start with the identity $$\eta(t):=\sum_{k=0}^{n-1}\frac1{t+\omega^k}=\frac{g'(t)}{g(t)}=\frac{nt^{n-1}}{t^n+1}. \quad\quad\quad\quad\quad\quad\quad\quad(1)$$ Your initial sum is $$2\sum_{j,k=0}^{n-1}\frac1{\omega^k+\omega^{-k}+\omega^j+\omega^{-j}}=2\sum_{j,k=0}^{n-1}\frac1{(\omega^k+\omega^{j})(1+\omega^{-j-k})}.$$ Fix $$j+k\equiv a\pmod n$$, and change the order of summation from $$(j,k)$$ to $$(a,k)$$. We get $$\sum_{a=0}^{n-1}\frac1{1+\omega^{-a}}\sum_{k=0}^{n-1}\frac2{\omega^k+\omega^{a-k}}.$$ Consider the inner sum. Denote $$t_{\pm}(a)=\pm i\omega^{a/2}$$ so that $$\frac2{\omega^k+\omega^{a-k}}=\frac1{\omega^k+t_+(a)}+\frac1{\omega^k+t_-(a)}$$ and by (1) we get $$\sum_{k=0}^{n-1}\frac2{\omega^k+\omega^{a-k}}=n\left(\frac{t_+(a)^{n-1}}{t_+(a)^n+1}+\frac{t_-(a)^{n-1}}{t_-(a)^n+1}\right)=nt_+(a)^{n-1}=n(-1)^{\frac{n-1}2}\omega^{\frac{n-1}2 a}.$$ So it remains to calculate the sum $$n(-1)^{\frac{n-1}2}\sum_{a=0}^{n-1} \frac{\omega^{\frac{n-1}2 a}}{1+\omega^{-a}}= n(-1)^{\frac{n-1}2}\sum_{z\in \Omega}\frac{z^{\frac{n+1}2}}{1+z}.$$ Denote $$m=\frac{n+1}2$$. We get $$\sum_{z\in \Omega} \frac{z^m}{z+1}= \sum_{z\in \Omega} \frac{z^m-(-1)^m+(-1)^m}{z+1}= (-1)^m\cdot \frac{n}2+\sum_{z\in \Omega} (z^{m-1}-z^{m-2}+\ldots+(-1)^{m-1})=\\ (-1)^m\cdot \frac{n}2+(-1)^{m-1}\cdot n=(-1)^{m-1}\cdot \frac{n}2$$ and your conjecture follows.