I look for possiblity to define sin/cos through algebraic relations without involving power series, integrals, differential equation and geometric intuition.

Is it possible to define sin and cos through some axioms?


$$\sin 0 = 0, \cos 0 = 1$$ $$\sin \pi/2 = 1, \cos \pi/2 = 0$$ $$\sin^2 x + \cos^2 x = 1$$ $$\sin(x+2\pi n) = \sin x, \cos(x+2\pi n) = \cos x$$ $$\sin(-x)=-\sin x, \cos(-x) = \cos x \text{ for } x \in [-\pi;0]$$ $$\sin(x+y)=\sin x \cos y + \sin y \cos x$$

and able to prove trigonometric school equations?

What additions required to prove continuity and uniqueness of such functions and analysis properties like:

$$\lim_{x \to 0}\frac{\sin x}{x} = 0$$ or $$\sin ' x = \cos x$$ or $$\int \frac{dx}{\sqrt {1-x^2}} = \arcsin x$$

PS In Walter Rudin book "Principles of Mathematical analysis" sin and cos introduced through power series.

In Solomon Feferman book "The Number Systems: Foundations of Algebra and Analysis" I see system derived from integral definition.

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    $\begingroup$ How would you define $\pi$? I mean, without geometry or trigonometric functions. $\endgroup$ – ajotatxe May 28 '15 at 20:05
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    $\begingroup$ You might be interested in this. I plan on expanding on it to define $\sin$ and $\cos$ for real numbers. $\endgroup$ – GPerez May 28 '15 at 20:08
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    $\begingroup$ I answer to myself: $\pi$ can be defined as $6\sqrt{\sum_{n=1}^\infty(1/n^2)}$. But trying to define exp, sin or cos from this deos not look nice... $\endgroup$ – ajotatxe May 28 '15 at 20:10
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    $\begingroup$ Relevant $\endgroup$ – Akiva Weinberger May 29 '15 at 0:30
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    $\begingroup$ There's this question, which gives a perfectly good definition. $\endgroup$ – Milo Brandt May 29 '15 at 1:40

To quote a previous answer of mine:

Robison, "A new approach to circular functions, π, and lim sin(x)/x", Math. Mag. 41.2 (March 1968), 66–70 [jstor].

In this paper it is shown that the addition law for cosine (and a couple other simple assumptions) uniquely determines cosine and sine.

(This is the paper I cite most often on StackExchange.)

If you don't have access to jstor (and don't want to sign up for their free 3-papers-at-a-time deal), you could try this other answer of mine to a closely related question about exponentiation, in which I adapted Robison's proof to give the following functional characterization of sine and cosine:

Proposition 1. Suppose $C,S\colon\mathbb R\to\mathbb R$ satisfy these conditions:

  1. $C$ and $S$ are continuous;
  2. $C(u-v) = C(u)C(v)+S(u)S(v)$ for all $u,v\in\mathbb R$;
  3. $S(u-v) = S(u)C(v)-C(u)S(v)$ for all $u,v\in\mathbb R$;
  4. $C$ and $S$ are not both identically zero.

Then there exists $\lambda\in\mathbb R$ such that

$$ C(u) = \cos(\lambda u) \quad\text{and}\quad S(u) = \sin(\lambda u) \text{ .} $$

You'll note that I just took continuity as an axiom, which was convenient and natural in the context of that other question; Robison proves continuity from weaker but less intuitive axioms — if memory serves, the key one is that cosine has a smallest positive root, i.e., there exists $p>0$ such that $C(p)=0$ and $C(x)\ne0$ for $0 < x < p$ — which might serve your purposes better.

(Oh, and since there's some question about degrees and radians and all that: Robison takes the normalization $p=1$, proves that $\lim_{x\to 0}\frac{S(x)}{x}$ exists, then defines $\pi = 2\lim_{x\to 0} \frac{S(x)}{x}$, and then defines $\sin(x) = S(\frac{2x}{\pi})$, and similarly for $\cos$. This manoeuvre amounts to defining $\pi$ by the condition $\lim_{x\to 0}\frac{\sin x}{x} = 1$, very much like we sometimes define $e$ by the condition $\lim_{x\to 0}\frac{e^x-1}{x} = 1$.)

  • $\begingroup$ I gave some more conditions that specify radians, if you like. (For some reason, I find inequalities more pleasing than limits. Perhaps because inequalities are more elementary of an idea.) $\endgroup$ – Akiva Weinberger May 29 '15 at 4:25
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    $\begingroup$ Yes, I saw your answer. Nice observations. You might also be interested to know that $f(x) = e^x$ is the only function satisfying $f(x+y)=f(x)f(y)$ and $f(x)\ge x+1$. $\endgroup$ – user21467 May 29 '15 at 4:28
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    $\begingroup$ $e^x\ge x+1$ is my favorite inequality, actually. It's surprisingly powerful — for example, it lets you prove that $\ln2=1-\frac12+\frac13-\dotsb$ without using calculus! (You need the squeeze theorem at the very end, though.) $\endgroup$ – Akiva Weinberger May 29 '15 at 4:30
  • $\begingroup$ Very nice answer! $\endgroup$ – Noah Schweber Jun 18 '15 at 16:53

From what you do not want that pretty much leaves functional equations.

There seems to be a system of two functional equations for sine and cosine: (link)

$$ \Theta(x+y)=\Theta(x)\Theta(y)-\Omega(x)\Omega(y) \\ \Omega(x+y)=\Theta(x)\Omega(y)+\Omega(x)\Theta(y) $$

  • $\begingroup$ @ajotatxe Through power series? $\endgroup$ – Arthur May 28 '15 at 20:06
  • $\begingroup$ Via $f(x+y) = f(x) f(y)$ plus some extra condition $f(0) = 1$? Not sure if that is enough for complex argument. Hm. $\endgroup$ – mvw May 28 '15 at 20:06
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    $\begingroup$ Since these formulas hold for angles measured in degrees or radians alike, could there still be some ambiguity? I.e., would this only allow us to know $\sin(kx)$ and $\cos(kx)$? $\endgroup$ – pjs36 May 28 '15 at 20:55
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    $\begingroup$ The functions $\Theta(x)=0$ and $\Omega(x)=0$ also satisfy the equations. $\endgroup$ – PyRulez May 29 '15 at 0:28
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    $\begingroup$ @PyRulez See StevenTaschuk's answer below, prop. 1 for the extra conditions. $\endgroup$ – mvw May 29 '15 at 0:32

A definition not involving power series, integrals, differential equations, or geometric intuition is:$$\cos x+\mathrm i\sin x=\lim_{n\rightarrow\infty}\left(1+\frac{\mathrm ix}{n}\right)^{\!n}\quad(x\in\Bbb R).$$

  • $\begingroup$ You can actually get that definition through geometric intuition. $\endgroup$ – Akiva Weinberger May 29 '15 at 0:31
  • $\begingroup$ @columbus8myhw: How? $\endgroup$ – cfh May 29 '15 at 8:05
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    $\begingroup$ @cfh Basically, because $\sin\epsilon\approx\epsilon$ and $\cos\epsilon\approx1$ for small $\epsilon$, we have $1+\frac{ix}n\approx\cos\frac x n+\sin\frac x n$. Using De Moivre, we get $(1+\frac{ix}n)^n\approx\cos x+\sin x$, and the approximation becomes equality taking the limit. $\endgroup$ – Akiva Weinberger May 29 '15 at 10:35
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    $\begingroup$ (First comment is "geometric" because this all has a nice geometric interpretation in the unit circle of the complex plane.) $\endgroup$ – Akiva Weinberger May 29 '15 at 10:41
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    $\begingroup$ @cfh $\sin\epsilon\approx\epsilon$ is basically saying that a small arc is well approximated by a line. $\epsilon$ is the length of the arc, and $\sin\epsilon$ is its height. The arc is almost entirely vertical, so… $\endgroup$ – Akiva Weinberger May 29 '15 at 12:38

Other answers give equations that define sine and cosine without specifying the units. If you want radians, you'll want: \begin{align} \sin x<x<\tan x&&&x\in\left(0,\frac\pi2\right) \end{align} There are other equivalent ways to get radians, such as: \begin{align} \sin x&\le x^2+x&x&\in\mathbb R\\ \cos x&<\frac{\sin x}x<1&x&\in(-\pi,\pi)-\{0\} \end{align}

As for getting $\pi$, here's one I just remembered: Let's say you have formulae that define sine and cosine without units. Well, you don't even need the right units to define $\pi$!

Call our functions $\sin x^\circ$ and $\cos x^\circ$ (even though they might not be degrees — the $^\circ$ just means "some unit" here). Let the smallest positive root of $\cos x^\circ$ be called $r$.

Well, $\frac\pi4$ is the unique number such that: \begin{align} \cos x^\circ&>\frac\pi4\left(1-\frac{x^2}{r^2}\right)&x&\in\mathbb R \end{align} Link to a helpful graph to help you see what's going on. Try changing $r$; you'll see the inequality always holds. Try changing $p$; you'll see that this inequality uniquely defines $\pi$.

  • $\begingroup$ Honestly, you're forced to use radians the minute you start thinking about trig inequalities. (Also: Did you know that the local maxima and minima of $\dfrac{\sin x}x$ are precisely at the intersections of the graphs of $\dfrac{\sin x}x$ and $\cos x$? Graph it. This only works for radians, again.) Similarly, the benefits of using $e^x$ for exponentials, with that weird base $e$, becomes apparent once you start dealing with inequalities. $\endgroup$ – Akiva Weinberger May 29 '15 at 1:15
  • $\begingroup$ Calculus is just inequalities disguised. The $\epsilon-\delta$ definition of a derivative is full of inequalities. $\endgroup$ – Akiva Weinberger May 29 '15 at 1:17
  • $\begingroup$ Note that the appearance of $\pi/2$ in the inequalities is not necessary for the axiomatization (in case one considers $\pi$ to be assuming the consequent) - one can replace it with $1$ or any sufficiently small positive $x$. $\endgroup$ – Mario Carneiro Jun 2 '15 at 22:44
  • $\begingroup$ @MarioCarneiro Right, yes. $\endgroup$ – Akiva Weinberger Jun 3 '15 at 0:21

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