Which properties characterize $\sin, \cos$? I know a few properties of $\sin$ and $\cos$, for example:


*

*$\sin^2+\cos^2=1$

*$\sin (a+b) = \sin a\cos b+\cos a\sin b$.

*$\cos (a+b) = \cos a\cos b-\sin a\sin b$.

*$\sin (x+\delta) = \sin x$ for some real $\delta$.

*$\cos (x+\delta) = \cos x$ for the same real $\delta$.


Obviously, the list is incomplete. 
My question is: Which set of properties characterize $\sin$ and $\cos$ (i.e, if $s,s',c,c'$ satisfy the properties, then $s=s', c=c'$) ? 
I'm looking for a set of such properties that doesn't reference differentiability, continuity, or the number $\pi$.

E: I've given up on the requirement of not assuming continuity (although, if someone does get a proof without assuming it, that person will get a nice bounty $:$-$)$ ), however, I still don't want differentiability or $\pi$ in this characterization.
Just to be clear, I'm looking for a proof of something like

The continuous solutions to $f^2+g^2=1$, $f(a+b)=f(a)g(b)+g(a)f(b)$ and $g(a+b)=g(a)g(b)-f(a)f(b)$ are unique.

 A: I am not sure whether this is what you're asking. Although the conditions do not reference differentiability, continuity or the number $\pi$,  it is used in the proof.

Theorem: If $f: \mathbb R \rightarrow \mathbb R,g : \mathbb R \rightarrow \mathbb R$ are functions that satisfy the following conditions:

*

*$f,g$ are periodic with some period $4\delta$ with $\delta \neq 0$.

*$0 \leq g(x) \leq x$ for all $0 \leq x \leq 2\delta$.

*$-x \leq g(x)\leq 0$ for all $2\delta \leq x \leq 4\delta$.

*$-x+1 \leq f(x) \leq 1$ for all $0 \leq x \leq 1$.

*$x+1 \leq f(x) \leq 1$ for all $x \leq 0$.

*$g(x+y)=g(x)f(y)+g(y)f(x)$ for all $x,y \in \mathbb R$.

*$f(x+y)+f(x-y)=2f(x)f(y)$ for all $x,y \in \mathbb R$.

*$g(x) \geq x \cdot f(x)$ for all $-1 \leq x \leq 1$.

*$f(\delta)=0$,  $g(\delta)=1$
then $f(x)=\cos(x)$ and $g(x)=\sin(x)$.

Part $(1)$, $(2)$ and $(3)$ imply that $g$ is bounded.
Part $(1)$, $(4)$ and $(5)$ imply that $f$ is bounded.
Part $(2)$ and $(3)$ imply that $\lim_{x \to 0} g(x) = 0$ and that $g(0)=0$.
Part $(4)$ and $(5)$ imply that $\lim_{x \to 0} f(x) = 1$ and that $f(0)=1$.
Part $(6)$ implies that \begin{align*} \lim_{x \to a} g(x) &= \lim_{x \to a} g(x-a)f(a)+g(a)f(x-a) \\ &=  \lim_{y \to 0} g(y)f(a)+g(a)f(y) \\  &= f(a) \lim_{y \to 0} g(y)+ g(a) \lim_{y \to 0} f(y) = g(a) \end{align*}
This implies that $g$ is continuous.
Condition $(6)$ and $(9)$ imply that $g(x+\delta)=g(\delta)f(x)=f(x)$, hence $f$ is continuous.
Condition $(7)$ is the d'Alembert functional equation. It is known that the only continuous functions $f$ satisfying it are
$f(x)=0, f(x)=1, f(x)=\cos(kx), f(x)=\cosh(kx)$ for some $k \in \mathbb R$. The only one that satisfies condition $(1)$ is $f(x)=\cos(kx)$. This implies that $g(x)=\sin(kx)$.
Condition $(2)$, $(3)$ and $(8)$ give $1 \geq \frac{g(x)}{x} \geq f(x)$ for $-1 \leq x \leq 1$ and this implies $g'(0) = \lim_{x \to 0} \frac{g(x)-g(0)}{x} = \lim_{x \to 0} \frac{g(x)}{x} = 1$ and hence $k=1$, so $f(x)=\cos(x)$ and $g(x)=\sin(x)$.
