TITLE: Axiomatic definition of sin and cos?
QUESTION [34 upvotes]: 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?
Like:
$$\sin 0 = 0, \cos 0 = 1$$
$$\sin \pi/2 = 1, \cos \pi/2 = 0$$
$$\sin^2 x + \cos^2 x = 1$$
$$\sin(x+2\pi n) = \sin x, \cos(x+2\pi n) = \cos x$$
$$\sin(-x)=-\sin x, \cos(-x) = \cos x \text{ for } x \in [-\pi;0]$$
$$\sin(x+y)=\sin x \cos y + \sin y \cos x$$
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.

REPLY [3 votes]: 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 $\pi$.