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Another question that's getting no answers on stackexchange:

Once upon a time, when Wikipedia was only three-and-a-half years old and most people didn't know what it was, the article titled functional equation gave the identity $$ \sin^2\theta+\cos^2\theta = 1 $$ as an example of a functional equation. In this edit in July 2004, my summary said "I think the example recently put here is really lousy, because it's essentially just an algebraic equation in two variables." (Then some subsequent edits I did the same day brought the article to this state, and much further development of the article has happened since then.)

The fact that it's really only an algebraic equation in two variables, $x^2+y^2=1$, makes it a lousy example of a functional equation. It doesn't really involve $x$ and $y$ as functions of $\theta$, since any other parametrization of the circle would have satisfied the same equation. In a sense, that explains why someone like Norman Wildberger can do all sorts of elaborate things with trigonometry without ever using trigonometric functions.

But some trigonometric identities do involve trigonometric functions, e.g. $$ \sin(\theta_1+\theta_2)=\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2 $$ $$ \sec(\theta_1+\cdots+\theta_n) = \frac{\sec\theta_1\cdots\sec\theta_n}{e_0-e_2+e_4-e_6+\cdots} $$ where $e_k$ is the $k$th-degree elementary symmetric polynomial in $\tan\theta_1,\ldots,\tan\theta_n$. These are good examples of satisfaction of functional equations.

So at this point I wonder whether all trigonometric identities that do seem to depend on which parametrization of the circle is chosen involve adding or subtracting the arguments and no other operations. In some cases the addition or subtraction is written as a condition on which the identity depends, e.g. \begin{align} \text{If } & x+y+z=\pi \\ \text{then } & \tan x+\tan y+\tan z = \tan x\tan y\tan z. \end{align}

QUESTION: Do all trigonometric identities that do involve trigonometric functions, in the sense that they are good examples of satisfaction of functional equations by trigonometric functions, get their non-triviality as such examples only from the addition or subtraction of arguments? Or is there some other kind? And if there is no other kind, can that be proved?

In comments below the stackexchange posting, Gerry Myerson mentioned the identities $$ \cos \frac x2=\sqrt{\frac{1+\cos x}{2}} $$ and $$ \prod_{k=1}^\infty \cos\left(\frac{x}{2^k}\right)= \frac{\sin x}{x} $$ The latter is somewhat like the one involving tangents above: One can say that if $x_n = x_{n+1}+x_{n+1}$ for $n=1,2,3,\ldots$ then $$ \prod_{k=1}^\infty \cos \left(\frac{x_k}{2^k}\right) = \frac{\sin x_1}{x_1}. $$ A similar but simpler thing applies to the half-angle formula.

Postscript: Wikipedia's list of trigonometic identities is more interesting reading than you might think. It has not only the routine stuff that you learned in 10th grade, but also some exotic things that probably most mathematicians don't know about. It was initially created in September 2001 by Axel Boldt, who was for more than a year the principal author of nearly all of Wikipedia's mathematics articles---several hundred of them.

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  • $\begingroup$ I'm a little confused: are you saying that $sin^2 θ + cos^2 θ = 1$ a bad example of a functional equation because $sin θ$ and $cos θ$ are not the only continuous solutions, in contrast to, say, $e^{x+y}=e^x e^y$? $\endgroup$
    – Paul Siegel
    Commented Jan 19, 2012 at 22:05
  • $\begingroup$ @Paul Siegel : Let $f(t) = (t^2 - 1)/(t^2 + 1)$ and $g(t) = 2t/(t^2 + 1)$. The $f(t)^2 + g(t)^2 = 1$. That's what I meant when I wrote "any other parametrization of the circle would have satisfied the same equation." $\endgroup$
    – Michael Hardy
    Commented Jan 19, 2012 at 22:41
  • $\begingroup$ In case anyone wants to see what Michael doesn't accept as an example, the m.se link is math.stackexchange.com/questions/99909/… $\endgroup$
    – Gerry Myerson
    Commented Jan 20, 2012 at 4:47
  • $\begingroup$ Robert Israel has suggested the identity $\cos(t \sin(x)) = J_0(t) + 2 \sum_{k=1}^\infty J_{2k}(t) \cos(2kx)$. If we fix $t=1$ that becomes $\cos(\sin(x)) = J_0(1) + 2 \sum_{k=1}^\infty J_{2k}(1) \cos(2kx)$ and then it doesn't have anything like the function $t\mapsto J_{2k}(t)$, which is moderately exotic for something bearing a "trigonometry" tag, but it does have $k\mapsto J_{2k}(t)$, which is similarly moderately exotic. This does seem like a counterexample to the simplest interpretation of my guess. But one could ask whether the guess is still true if..... $\endgroup$
    – Michael Hardy
    Commented Jan 20, 2012 at 19:26
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    $\begingroup$ What about differential equations like y''+y=0? $\endgroup$
    – Anixx
    Commented Oct 9, 2020 at 16:46

2 Answers 2

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Most of these identities have a different explanation, via holonomic functions, aka solutions of linear differential equations with polynomial coefficients. And, in the end, it is all a matter of linear algebra and bounding dimensions. This is very well explained in slides by Bruno Salvy (where trigonometrics are explicitly used for the simple examples, before moving on to the 'fun' stuff).

To a certain extent, there is still a lot of geometry still going on in the above, but the geometry of the parameter space, rather than looking directly at the traces of the functions.

Many details of the method can be found in the paper A non-holonomic systems approach to special function identities by Chyzak, Kauers and Salvy.

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Unfortunately the category of smooth manifolds is not closed under finite limits. (Yes, to annoy my Wikipedian colleague Michael, I'm speaking Bourbaki.) This kind of undermines the idea of setting up an actual answer in terms of taking the usual map from the line to the circle, qua morphism, and generating some "category of relations" within which the question has a chance of an acceptable answer. But perhaps there is a germ of an idea here. Formulate the question via some equational logic that admits the more general parametrisations of the circle.

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