About the addition formula $f(x+y) = f(x)g(y)+f(y)g(x)$ Consider the functional equation
$$f(x+y) = f(x)g(y)+f(y)g(x)$$
valid for all complex $x,y$. The only solutions I know for this equation are $f(x)=0$, $f(x)=Cx$, $f(x)=C\sin(x)$ and $f(x)=C\sinh(x)$.
Question $1)$ Are there any other solutions ?

If we set $x=y$ we can conclude that if there exists a $g$ for a given $f$ then $g$ must be $g(x)=\frac{f(2x)}{2f(x)}$. By using this result, I tried setting $y=2x$ yielding
$$f(x+2x)=f(x)g(2x)+g(x)f(2x)=\dfrac{f(x)f(4x)}{2f(2x)}+\dfrac{f^2(2x)}{2f(x)}$$
Question $2)$ Are $f(x)=0$,$f(x)=Cx$,$f(x)=C\sin(x)$ and $f(x)=C\sinh(x)$ the only solutions to $f(3x)=\dfrac{f(x)f(4x)}{2f(2x)}+\dfrac{f^2(2x)}{2f(x)}$? If not, what are the other solutions?

As leshik pointed out, this equation has plenty of discontinuous solutions (e.g. for $g=1$ it becomes Cauchy's functional equation), so let's just consider continuous solutions. $f=0$ is the trivial solution; from now on we will assume that $f$ is not identically zero. We consider two cases:


*

*If $f(x)$ and $g(x)$ are linearly dependent, then there is some $\lambda \neq 0$ such that $g(x)=\lambda f(x)$. Since we can replace $f(x)$ with $c f(x)$ and the original equation is still satisfied, we may assume without loss of generality that $g(x)=\frac{1}{2}f(x)$, so the equation becomes $f(x+y)=f(x)f(y)$. Now since $f(x+0)=f(x)f(0)$ and $f$ is not identically zero, we must have $f(0) \neq 0$. Then since $f(0)=f(0+0)=f(0)^2$, $f(0)=1$. By continuity $F(t) = \ln (2f(t))$ is defined in a neighborhood of $0$ and satisfies $F(x+y)=F(x)+F(y)$. This is Cauchy's functional equation, so since $F$ is continuous $F(x)=rx$, and therefore $f(x)=e^{rx}$ and $g(x)=\frac{1}{2}e^{rx}$. Therefore the general solution in this case is:


*

*$f(x)=ke^{rx}$, $g(x)=\frac{1}{2}e^{rx}$.


*Now suppose that $f(x)$ and $g(x)$ are linearly independent. Since $f(0)=f(0+0)=2f(0)g(0)$, we either have $f(0)=0$, or else $f(0) \neq 0$ and $g(0)=\frac{1}{2}$. In the latter case we would have $f(x)=f(x+0)=\frac{1}{2}f(x)+f(0)g(x)$, so $f(x)$ and $g(x)$ would be linearly dependent. Therefore $f(0)=0$.
Since $f$ is not identically zero, there is no $x$ for which both $f(x)$ and $g(x)$ are zero, since otherwise $f(x+y)=f(x)g(y)+f(y)g(x)$ would be identically zero. Then by continuity there must be some $p$ such that $f(p) \neq 0$ and $g(p) \neq 0$. It follows that $g(x)=\frac{1}{f(p)}f(x+p)-\frac{g(p)}{f(p)}f(x)$, so $g$ is a linear combination of $f$ and its translate $f_p$ (where $f_p(x)=f(x+p))$. Since $f$ and $g$ are linearly independent, it follows that $f$ and $f_p$ are linearly independent. Then since $f(x+y)=f(x)g(y)+f(y)g(x)$ for all $x,y$, every translate $f_y$ of $f$ is a unique linear combination of $f$ and $g$, and therefore every translate $f_y$ of $f$ is a unique linear combination of $f$ and $f_p$.
Since all translates of $f$ are linearly combinations of the linearly independent functions $f$ and $f_p$, this implies that $f$ is differentiable as described in the answer to this question. Then by the answers to this question, $f$ is a solution to the ODE $\frac{d^{2}f}{dx^{2}}+B\frac{df}{dx}+Cf=0$ for some $B,C \in \mathbb{R}$, other than the solution $ce^{rx}$. With the boundary conditions $f(0)=0$ it follows that there are only three families of solutions:


*

*$f(x) = kxe^{rx}$, $g(x)=e^{rx}$

*$f(x) = ke^{rx}\sin(sx)$, $g(x)=e^{rx}\cos(sx)$

*$f(x)= ke^{rx}\sinh(sx)$, $g(x)=e^{rx}\cosh(sx)$
Of course, these are valid solutions because of the following addition formulae 
$$(x+y)=x\cdot 1 + y \cdot 1$$
$$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$$
$$\sinh(x+y)=\sinh(x)\cosh(y)+\sinh(y)\cosh(x)$$
We have found all the continuous solutions of the given functional equation. QED
