Characterizations of cosine and sine functions

Question

Does the following set of rules characterize cosine and sine functions?

1. $C(x)$ and $S(x)$ are $2\pi$-periodic, with $2\pi$ the smallest period.

2. $C(x)$ is even and $S(x)$ is odd.

3. $C(0)=1, S(0)=0$.

4. $C(x+y)=C(x)C(y)-S(x)S(y)$; $S(x+y)=S(x)C(y)+S(y)C(x)$.

If so, are any of these rules redundant? Where can I find a proof?

Answer 1

I would add $C(x)^2+S(x)^2=1$.

Moreover, those that you wrote are the main properties of sine\cosine. Who assures that $C(\pi)=-1$?

Those, instead, are properties fro a periodic function, with given initial point, and with a link to another periodic function, but I wouldn't say those coincide with sine\cosine (which raise from geometry).

Answer 2

The constant functions $C(x)=1$ and $S(x)=0$ satisfy these conditions.

Now that you added the condition of periodicity exactly $2\pi$, the only continuous spurious solution is $C(x)=\cos(x)$ and $S(x)=-\sin(x)$. However there are still plenty of other discontinuous solutions.

Here is why. First note that if one would have $C(x)=S(x)=0$ for any particular $x$, condition 4 with $y=-x$ would contradict condition 3, so this does not happen. So one can define a new function $\def\R{\mathbf R}\def\C{\mathbf C}f:\R\to\C^\times$ by $\def\i{\mathbf i}f(x)=C(x)+\i S(x)$, and the given conditions amount to saying that $f$ is a group morphism (condition 4), for which $\def\Z{\mathbf Z}\ker(f)=2\pi\Z$ (condition 1), and such that $f(-x)=\overline{f(x)}$ for all $x$ (condition 2). Since also $f(x)f(-x)=1$ for all $x$, the latter condition amounts to $|f(x)|=1$ for all $x$; we are looking essentially at an injective group morphism from $\R/2\pi\Z$ to the circle group $U(1)\subset\C^\times$ which is of course isomorphic to $\R/2\pi\Z$ itself. The theory of representations of the circle group tells us that all the continuous group morphsims $U(1)\to\C^\times$ have their image in $U(1)$ and are in fact all given by $z\mapsto z^k$ for some $k\in\Z$ (of which only $k=1,-1$ give injective morphisms).

However without continuity, it is the abstract structure of the group $U(1)$ that matters, and it turns out to be a direct sum of its torsion subgroup (the roots of unity) isomorphic to $\def\Q{\mathbf Q}\Q/\Z$, and a second factor that is an uncountable sum of copies of the additive group$~\Q$. The second factor has many injective endomorphisms (any injective $\Q$-linear map will do). I believe the torsion subgroup $\Q/\Z$ has many automorphisms as well, since it decomposes as direct sum of it $p$-torsion subgroups, the Prüfer $p$-groups, and each of these has plenty of automorphisms. The latter point means you cannot expect to deduce from the given condition even the values of $f(x)$ for all $x\in\pi\Q$ (although you can show that $f(\pi)=-1$: one has $f(\pi)^2=1$, and if $f(\pi)$ were $1$ then $f$ would be $\pi$-periodic).