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

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.)