Seeking elegant proof of $\sum_{cyc}\cot(x-y)\cot(y-z)=1$

Question

Problem

The following identity is obvious: $$ \frac{1}{(x-y)(y-z)} + \frac{1}{(y-z)(z-x)} + \frac{1}{(z-x)(x-y)} = 0 $$

This post is for a trigonometric version in terms of cotangent: $$ \cot (x- y) \cot( y- z) + \cot( y-z) \cot( z - x) + \cot ( z- x) \cot ( x-y) = 1 $$

The following Mathematica code gives me 1

Simplify[Cot[x - y] Cot[y - z] + Cot[z - x] Cot[y - z] +  Cot[x - y] Cot[z - x]]

So I think the identity is right.

I could have proved it using a brute force expansion in terms of trigonometric functions of $x, y,z$ individually. But I feel there definitely is a more elegant proof. I'm looking for such a proof.

Follow-up question: is there a similar identity involving the elliptic function sn, cn, dn? I don't know the answer, but it is possible based on the context of the real world problem.

Thanks!

Answer 1

Given any three numbers $\;x,y,z\;$ define $$ X := e^{ix},\; Y := e^{iy},\; := e^{iz}\quad \text{ and }\quad a := X^2,\; b := Y^2,\; c := Z^2. $$ Then, by definition of cotangent $\; \cot(x-y) = i(a+b)/(a-b).\;$ Thus $$ \cot(x-y)\cot(y-z) = -\frac{(a+b)(b+c)}{(a-b)(b-c)}. $$ Cycle, add, and use common denominator to get $$ 1 - \sum_{cyc}\cot(x-y)\cot(y-z) = \frac{S}{ (a-b)(b-c)(c-a)} $$ where $$ S := (a\!-\!b)(b\!-\!c)(c\!-\!a) \!+\! (a\!+\!b)(b\!+\!c)(c\!-\!a)\!+\! (a\!-\!b)(b\!+\!c)(c\!+\!a)\!+\! (a\!+\!b)(b\!-\!c)(c\!+\!a) .$$ Now $\;S\;$ expands to identically zero. There are several ways to verify this. For example, if $$ T(b) := (a-b)(b-c)+(a+b)(b+c) = (b+b)(c+a),$$ (which is equivalent to id$_{3,3,1,2e}$) then $S = T(b)(c-a) + T(-b)(c+a), $ but $ T(b)=-T(-b). $ Thus, $\;0 = S\;$ is the identity named as id$_{3,4,1,3b}$ in my collection of hundreds of Special Algebraic Identities. This answers your main question.

You also asked

Follow-up question: is there a similar identity involving the elliptic function sn, cn, dn? I don't know the answer, but it is possible based on the context of the real world problem.

Depends on what precisely you mean by "similar identity". I suggest that you look at my identities for those tagged [JE] just as id$_{3,4,1,3b}$. That is, this identity is already satisfied by Jacobi sn, sc, sd. More precisely, suppose the function $\;f\;$ is any one of $\;\sin,\;\tan,\;$sn, sc, sd. Then $$ 0 = f(a-b)f(b-c)f(c-a) + f(a+b)f(b+c)f(c-a) + \\ f(a-b)f(b+c)f(c+a) + f(a+b)f(b-c)f(c+a). $$

There are several other identities labeled [JE] that all have the same solutions as this identity.

This is related to MSE question 4088531 about functional equations for both sine and tangent.