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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 The Wiki link is a treasure chest of new questions. –  draks ... Jan 17 '12 at 18:44
+1 Very interesting question that has never occurred to me before. –  Alex Becker Jan 17 '12 at 19:21
If we're going to have "identities" with infinitely many terms, how about $$\cos(t \sin(x)) = J_0(t) + 2 \sum_{k=1}^\infty J_{2k}(t) \cos(2kx)$$? –  Robert Israel Jan 20 '12 at 9:01
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. –  Américo Tavares May 7 '12 at 18:34
It comes via Fourier series from the integral representation of the Bessel functions, $$ J_n(t) = \dfrac{1}{2\pi} \int_{-\pi}^\pi \exp(i n x - i t \sin(x))\; dx$$ –  Robert Israel Jan 2 at 1:05

2 Answers 2

$$\sin^2\theta+\cos^2\theta = 1$$ can be rewritten as functional equation: $$\sin^2\theta+\sin^2(\theta+\pi/2) = 1.$$

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But maybe the "trigonometric identity involving trigonometric functions" in this instance is $\cos\theta=\sin(\theta+\pi/2)$, without the Pythagorean identity. –  Michael Hardy May 13 at 15:58
But $\cos\theta=\sin(\theta+\pi/2)$ is very trivial and uninteresting. The relation of trigonometric functions with the Pythagorean identity is interesting. –  Martín-Blas Pérez Pinilla May 14 at 10:09
@MichaelHardy, interesting but paywalled: universitypublishingonline.org/maa/…. –  Martín-Blas Pérez Pinilla May 14 at 17:50

All the trigonometric identities can be derived from the properties of complex number multiplication.

The point on the unit circle at angle t in radians, as measured from the positive $x$-axis is $(\cos(t) + i \sin(t))$

The distance of this point from the origin is 1 because it is on the unit circle. Thus the magnitude of $( \cos(t) + i \sin(t) )$ is $1$ and this leads immediately to the identity

$$( \cos(t))^2 + ( \sin(t))^2 = 1$$

The sum of angle formulas follow immediately from the rule that to multiply two complex numbers you add their angles and multiply their magnitudes.

$$( \cos(t_1) + i \sin(t_1) ) \cdot ( \cos(t_2) + i \sin(t_2))$$

$$= \cos(t_1 + t_2) + i \sin (t_1 + t_2) (\cos (t_1) \cos(t_2) - \sin(t_1 ) \sin(t_2) ) + i ( \cos(t_1) \sin(t_2) + \sin(t_1) \cos(t_2) ) = \cos( t_1 + t_2) + i \sin(t_1 + t_2)$$

Separating real and imaginary components,

$$\cos(t_1) \cos(t_2) - \sin(t_1) \sin(t_2) = \cos(t_1 + t_2)$$

$$\cos(t_1) \sin(t_2) + \sin(t_1) \cos(t_2) = \sin(t_1 + t_2)$$

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Well, all this is standard stuff that I've known since I was in high school, but it doesn't really answer the question actually asked. –  Michael Hardy Jun 19 '12 at 1:22
This website accepts (prefers) math written in TeX. If you type \cos t and put dollar signs around it, it comes out $\cos t$, which looks better than cos(t). –  Gerry Myerson Jun 19 '12 at 3:57
I've edited this posting to set it in $\TeX$. –  Michael Hardy Jun 19 '12 at 18:11

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