Axiomatic definition of sin and cos? 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 be able to prove trigonometric school equations?
What additions are 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.
 A: A definition not involving power series, integrals, differential equations, or geometric intuition is:$$\cos x+\mathrm i\sin  x=\lim_{n\rightarrow\infty}\left(1+\frac{\mathrm ix}{n}\right)^{\!n}\quad(x\in\Bbb R).$$ 
A: 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:
  
  
*
  
*$C$ and $S$ are continuous;
  
*$C(u-v) = C(u)C(v)+S(u)S(v)$ for all $u,v\in\mathbb R$;
  
*$S(u-v) = S(u)C(v)-C(u)S(v)$ for all $u,v\in\mathbb R$;
  
*$C$ 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 $\lim_{x\to 0}\frac{e^x-1}{x} = 1$.)
A: 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 $\pi$.
A: 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)
$$
\Theta(x+y)=\Theta(x)\Theta(y)-\Omega(x)\Omega(y) \\
\Omega(x+y)=\Theta(x)\Omega(y)+\Omega(x)\Theta(y)
$$
