# Which trigonometric identities involve trigonometric functions?

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.

• +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
• @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.) – Michael Hardy Jan 20 '12 at 19:19
• 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 '14 at 1:05
• Ah, here it is: Abramowitz & Stegun formula 9.1.42. – Robert Israel Jan 2 '14 at 1:10

## 3 Answers

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,

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

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

• 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

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

If I understood correctly, the question is how uniquely we can determine trigonometric functions using functional equations. First thing is together with a strong enough functional equation we always need a some sort of "smoothness" condition. For example consider the system $$\begin{cases} f(x+y)=f(x)g(y)+g(x)f(y) \\ g(x+y)=g(x)g(y)-f(x)f(y) \end{cases}$$

Under any assumption sufficient for Cauchy functional equation we can deduce that it has a solution involving sine-and-cosine, but there are wildly discontinuous solutions one can construct using a Hamel bases of rationals over reals.

Next, since the set of all interesting functional equations is too big to consider, we can focus on somewhat special class called algebraic Addition theorems. For example $$h(x+y)=h(x)\sqrt{1-(h(y))^2}+\sqrt{1-(h(x))^2}h(y).$$ It is a theorem by Weierstrass that characterize all such meromorphic functions as either rational or periodic depending on the singularity type of $$h$$ at $$\infty.$$ You can see here for more details. Since the Jacobi's impossibility theorem of triply periodic meromorphic functions, we can find all of them explicitly. Roughly these are:

• trigonometric functions (including hyperbolic, and exponential functions)
• elliptic functions (I think of theses as some sort of generalized trigonometric functions, and I am not completely wrong as the theory of Jacobi elliptic functions can develop entirely analogous to trigonometry)
• rational functions (in particular polynomials).

So yes!, under some reasonable assumptions, algebraic addition theorems recover a generalized class of trigonometric functions. Here being meromorphic is important otherwise we should include algebraic functions (and possibly more!). Alternatively, I will roughly sketch a proof to convince you that any complex differentiable function with an algebraic addition theorem is the inverse of a certain kind of integral. Let $$\varphi$$ be an analytic function (on some domain of $$\mathbb{C}$$) with an addition theorem, then there is a polynomial $$G$$ with constant coefficients satisfying $$G(\zeta, \eta, \theta) =0,\tag1$$ where $$\zeta=\varphi(z), \eta=\varphi(w)=$$ and $$\theta=\varphi(w+z).$$ Since $$\zeta$$ and $$\eta$$ are independent variables, by considering total derivatives of $$G$$ with respect to them gives: $$\dfrac{\partial G}{\partial \zeta}\dfrac{d \zeta}{d z}+\dfrac{\partial G}{\partial \theta}\dfrac{d \theta}{d z}=\dfrac{\partial G}{\partial \eta}\dfrac{d \eta}{d w}+\dfrac{\partial G}{\partial \theta}\dfrac{d \theta}{d w}=0.$$ Also we know that $$\dfrac{d \theta}{d z}=\dfrac{d \theta}{d w},$$ and consequently $$\dfrac{\partial G}{\partial \zeta}\dfrac{d \zeta}{d z}-\dfrac{\partial G}{\partial \eta}\dfrac{d \eta}{d w}=0.\tag2$$ If $$\theta$$ does not appear in $$\dfrac{\partial G}{\partial \zeta}, \dfrac{\partial G}{\partial \eta},$$ this lase equation is an expression that contains only $$\zeta, \dfrac{d \zeta}{d z}, \eta$$ and $$\dfrac{d \eta}{d w}.$$ On the other hand, even if they contain $$\theta$$ we can eliminate it using both $$(1), (2)$$ to obtain a polynomial relation with constant coefficients $$H\left(\zeta, \dfrac{d \zeta}{d z}, \eta, \dfrac{d \eta}{d w}\right)=0.$$ Since this is true for all $$z, w$$ in the domain, we can evaluate $$v$$ at some specific point to obtain a first order differential equation $$\tilde{H}\left(\zeta, \dfrac{d \zeta}{d z}\right)=0.$$ If this evaluation of $$H$$ vanish in all its coefficient, we can try different values of $$w$$ until we find a non-zero equation $$\tilde{H}.$$ We can solve this equation in terms of an algebraic function $$K$$ such that $$\dfrac{d \zeta}{d z}=K(\zeta),$$ and hence after separating variables $$z-z_0=\displaystyle\int_{\zeta_0}^{\zeta}\dfrac{d\zeta}{K(\zeta)}.$$ where $$\varphi(z_0)=\zeta_0$$ are appropriate constants. Then the addition thereom of $$\varphi$$ can be rewritten as $$\displaystyle\int_{\varphi(z_0)}^{\zeta}\dfrac{d\zeta}{K(\zeta)}+\displaystyle\int_{\varphi(z_0)}^{\eta}\dfrac{d\zeta}{K(\zeta)}=\displaystyle\int_{\varphi(z_0)}^{\theta}\dfrac{d\zeta}{K(\zeta)}.$$

You can find the original argument of this algorithm in this book with all the details and much more. For example to what I described here: $$\begin{array}{|l|c|r|} \zeta=\varphi(z) & \theta & \tilde{H} \\ \hline \sin z & \zeta\sqrt{1-\eta^2}+\eta\sqrt{1-\zeta^2} & \left(\dfrac{d \zeta}{d z}\right)^2+\zeta^2-1 \\ \tan z & \dfrac{\zeta+\eta}{1-\zeta\eta} & \dfrac{d \zeta}{d z}-\zeta^2-1\\ \sec z & \dfrac{\zeta\eta}{1-\sqrt{1-\zeta^2}\sqrt{1-\eta^2}} & \left(\dfrac{d \zeta}{d z}\right)^2-\zeta^2(\zeta^2-1) \end{array}$$

I am too lazy to continue this for other trigonometric and hyperbolic functions. In the next simplest case $$K(\zeta)=K(\zeta, \sqrt{P(\zeta)})$$ is a rational function, where $$P$$ is a polynomial of degree $$3$$ or $$4$$ with no repeated roots, we recover the Jacobian elliptic functions. Similar computations can carry on for some other elliptic functions, in particular for lemniscate functions mentioned in this solution to one of your question.

• Excellent answer. – Piquito Jan 7 at 13:52