Defining sine and cosine from functional equations $(S(x))^2+(C(x))^2=1$, $C(x+y)=C(x)C(y)-S(x)S(y)$, etc

Question

Let $\text{S}(x)$ and $\text{C}(x)$ be non-constant, continuous, real functions such that the following properties are true.

(i) $\,\,\,\,(\text{S}(x))^2+(\text{C}(x))^2=1$

(ii)$\,\,\,\,\text{C}(x+y)=\text{C}(x)\text{C}(y)-\text{S}(x)\text{S}(y)$

(iii) $\,\,\text{S}(x+y)=\text{S}(x)\text{C}(y) +\text{C}(x)\text{S}(y)$

(iv) $\,\,\,\text{S}(0)=0$

(v)$\,\,\,\,\,\text{C}(0)= 1$

Do these properties uniquely define $\text{S}$ and $\text{C}$? (specifically do they define $\sin$ and $\cos$--although that would follow if $\text{S}$ and $\text{C}$ are uniquely defined)

If these three properties are not enough, can we add periodic properties to make them unique?

i.e.

$\text{(vi)}\,\,\,\,\,\,\text{S}(x+2\pi)=\text{S}(x)\,\,\,\,$

$\text{(vii)}\,\,\,\,\text{C}(x+2\pi)=\text{C}(x)$

This is similar to Defining sine and cosine but without using limit properties and adding (i) and potentially (iv) and (v).

What properties can we add that are not from analysis to make them unique?

Answer 1

Without continuity or nonconstantness assumption on $C$ and $S$, I would like to note that the conditions (ii) and (iii) already imply that either both $C$ and $S$ are identically zero, or (i), (iv), and (v) hold. From now on, we suppose that $C$ and $S$ obey both (ii) and (iii), and that they are nonconstant.

We can show that, for some (nonconstant) function $f:\mathbb{R}\to\mathbb{R}$ such that $$f(x+y)=f(x)+f(y)\text{ for any }x,y\in\mathbb{R}\,,\tag{*}$$ we have $$C(x)=\cos\big(f(x)\big)\text{ and }S(x)=\sin\big(f(x)\big)$$ for every $x\in\mathbb{R}$. Equation (*) is known as Cauchy's functional equation, and there are nontrivial functions $f$ (which are not continuous) that satisfy (*).

If, in addition, either $C$ or $S$ is continuous, then both are continuous and there exists $\lambda\in\mathbb{R}\setminus\{0\}$ such that $$f(x)=\lambda\, x$$ for all $x\in\mathbb{R}$. Hence, (ii) and (iii) along with continuity of $C$ and $S$, as well as nonconstantness of $C$ and $S$, show that there exists $\lambda\in\mathbb{R}\setminus\{0\}$ such that $$C(x)=\cos(\lambda \,x)\text{ and }S(x)=\sin(\lambda\, x)$$ for every $x\in \mathbb{R}$. (Notice that, when $\lambda =0$, we get another constant solution: $C(x)=1$ and $S(x)=0$ for all $x\in\mathbb{R}$.)

If you further assume (along with (ii), (iii), continuity, and nonconstantness) that either $C$ or $S$ is periodic with minimal period $P$, then both are periodic with minimal period $P\in\mathbb{R}_{>0}$, then $$C\left(\frac{P}{4}\right)=0\text{ and }S\left(\frac{P}{4}\right)=\pm 1\,.$$ If $s:=\sin\left(\dfrac{P}{4}\right)$, then $s\in\{-1,+1\}$ and $\lambda=s\dfrac{2\pi}{P}$, so $$C(x)=\cos\left(\frac{2\pi}{P}\,x\right)\text{ and }S(x)=s\,\sin\left(\frac{2\pi}{P}\,x\right)$$ for all $x\in\mathbb{R}$.

Particularly when $P=2\pi$, you get $$C(x)=\cos(x)\text{ and }S(x)=s\,\sin(x)$$ for each $x\in\mathbb{R}$. Unless there are other required properties, this is the best you can do with the conditions you put in your questions. For example, if you also require that $S(x)>0$ for every sufficiently small $x>0$, then you will get $s=+1$.

The user The_Sympathizer suggests (in a deleted answer) adding the condition $$\lim_{x\to 0}\,\frac{S(x)}{x}=1.$$ In this case, all we need are (ii), (iii), and the equation above in order to get that $C(x)=\cos(x)$ and $S(x)=\sin(x)$ for all $x\in \mathbb{R}$. (This is a great suggestion by the way.)

Answer 2

You also need to consider translations. If you translate $S$ and $C$ by any number $t$, the resulting functions have the same properties. So you need initial conditions such as : $S(0) = 0$ and $C(0) = 1$.