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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. $$ \text{If }x+y+z=\pi\text{ then }\tan x+\tan y+\tan z = \tan x\tan y\tan z. $$

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?

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.

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    +1 Very interesting question that has never occurred to me before.2012-01-17
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    Do you count half-angle formulas ($\cos(x/2)=\sqrt{(1+\cos x)/2}$) as addition/subtraction of arguments?2012-01-17
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    @GerryMyerson : I think one would have to count it that way. It's the same as $\cos w = \sqrt{(1+\cos(w+w))/2}$.2012-01-18
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    OK, then, how about $\prod_{k=1}^{\infty}\cos(x/2^k)=(\sin x)/x$?2012-01-18
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    @GerryMyerson : $x_2+x_2=x_1$, $x_3+x_3=x_2$, etc., and then $\prod_{k=1}^\infty \cos(x_k/2) = (\sin x)/x$.2012-01-18
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    If we're going to have "identities" with infinitely many terms, how about $$\cos(t \sin(x)) = J_0(t) + 2 \sum_{k=1}^\infty J_{2k}(t) \cos(2kx)$$?2012-01-20
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    @RobertIsrael : This looks like a counterexample to my guess. So a question is whether any such counterexample needs to have infinitely many terms? Or whether it must have something complicated like $k\mapsto J_{2k}(t)$? (One needn't go into $t\mapsto J_{2k}(t)$ since one would have a separate identity for each value of $t$ separately.)2012-01-20
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    PS: I see no reason not to allow infinitely many terms. At least not yet.2012-01-20
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    PPS: The aspect of it that qualifies it as a counterexample, at least to the simplest interpretation of my guess, is the left side---definitely not the right side. The right side is innocent.2012-01-20
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    If we allow angles and lengths of the sides we have in a triangle e.g. the law of sines.2012-05-07
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    @AméricoTavares : But that involves adding, i.e. the sum of the three angles is $\pi$, just as in the case I mentioned that says the sum of the tangents equals the product of the tangents if the sum of the three angles is $\pi$.2012-05-07
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    My rather trivial point is that instead of having an identity in the trigonometric functions of angles only (as in $\sin^2\theta+\cos^2\theta = 1$ or in $\sin(\theta_1+\theta_2)=\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2$) we also have the lengths of sides.2012-05-07
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    I'm not sure lengths of sides add a lot, unless you're dealing with polygons not inscribable in a sphere. The law of cosines says $a^2+b^2-2ab\cos C = c^2$. But if the circumscribed circle has unit diameter, then $a$, $b$, and $c$ are respectively $\sin A$, $\sin B$, and $\sin C$, and the identity says that if $A+B+C=\pi$ then $\sin^2 A+\sin^2 B - 2\sin A\sin B\cos C = \sin^2 C$.2012-05-07
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    Actually it occurs to me that the law of sines would still have some substance: it would say that if the circumscribed circle has unit diameter, then $a$, $b$, and $c$ are respectively $\sin A$, $\sin B$, and $\sin C$.2012-05-07
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    @AméricoTavares : "not inscribable in a sphere". Sorry---I meant not inscribable in a circle.2012-05-08
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    @MichaelHardy You are right.2012-05-08
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    How would you classify these? (It's Daviddaved@Wiki, thanks 4 your help there):http://math.stackexchange.com/questions/389225/a-matrix-w-integer-eigenvalues-and-trigonometric-identity2013-05-16
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    @Daved : Hello. At first glance that looks to me like something that doesn't depend on the fact that the circle is parametrized by arc length rather than in some other way.2013-05-16
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    @RobertIsrael : Months ago I added your identity $\cos(t \sin(x))$ $= J_0(t) + 2 \sum_{k=1}^\infty$ $J_{2k}(t) \cos(2kx)$ to Wikipedia's "list of trigonometric identities", and now I'm being asked if I can cite a source for it. Any ideas?2014-01-01
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    It comes via Fourier series from the integral representation of the Bessel functions, $$ J_n(t) = \dfrac{1}{2\pi} \int_{-\pi}^\pi \exp(i n x - i t \sin(x))\; dx$$2014-01-02
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    Ah, here it is: Abramowitz & Stegun formula 9.1.42.2014-01-02
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    @RobertIsrael : Perfect! Thank you.2014-01-02

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