0

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!

Ѕᴀᴀᴅ
  • 35,369
anecdote
  • 1,053
  • Mathematica's trigonometric algebra shows that it is correct. "A brute force expansion" in the post means that I can do TrigExpand for both Cos[x - y] Cos[y - z] Sin[z - x] + Cos[x - y] Sin[y - z] Cos[z - x] + Sin[x - y] Cos[y - z] Cos[z - x] and Sin[x - y] Sin[y - z] Sin[z - x] in terms of products like Cos[x] Cos[y]^2 Sin[x] etc. Again I checked in Mathematica, they are identical. What would be my proof. But I admit that I didn't check manually on paper. – anecdote Dec 07 '22 at 01:10
  • This is equivalent to $$\tan(x-y)+\tan(y-z)+\tan(z-x)=\tan(x-y)\cdot\tan(y-z)\cdot\tan(z-x).$$ – mr_e_man Dec 07 '22 at 01:21
  • it follows from something like $\tan(a+b) = \dfrac{\tan a + \tan b}{1-\tan a\tan b}$ and letting $c = -a-b$ and make an appropriate $x,y,z$ substitution. – dezdichado Dec 07 '22 at 02:00

1 Answers1

2

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.

Somos
  • 37,457
  • 3
  • 35
  • 85
  • It is indeed a proof. One converts a trigonometric identity into algebraic identities of the exponential. I up voted the answer. But the proof is done by an explicit computation. I will wait to see other proofs. The "elegant proof" I am looking, I hope, can provide guidance to the case of the elliptic functions. – anecdote Dec 07 '22 at 19:13
  • @anecdote What "guidance"? My equation is already the elliptic function case. – Somos Dec 07 '22 at 20:05