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?
 A: Some partial results:
Assume henceforth that neither $f$ nor $g$ is constant and $0$.
Take $y=0$, we get $$f(x)=f(x)g(0)+f(0)g(x),\textrm{ or }f(x)(1-g(0))=f(0)g(x)$$
 then either $g(0)=1$ and $f(0)=0$, or 

$$f(x)=\frac{f(0)}{1-g(0)}g(x)$$
  Substituting $x=0$ gives $g(0)=1-g(0)$ so $g(0)=0.5$, and hence $f(x)=2f(0)g(x)$.  Substituting into the original gives $$2f(0)g(x+y)=2f(0)g(x)g(y)+2f(0)g(y)g(x)$$ whence $$g(x+y)=2g(x)g(y)$$ As Kunnysan points out, $g(x)=e^x/2$ satisfies this.  Taking $x=n-1, y=1$ lets us prove inductively that $g(n)=2^{n-1}g(1)^n$.

Suppose now that $f(0)=0$ and $g(0)=1$.
Now, taking $y=-x$, we have  $$0=f(0)=f(x)g(-x)+f(-x)g(x)$$  Rearranging, we have (provided both expressions exist)$$\frac{f(x)}{f(-x)}=-\frac{g(x)}{g(-x)}$$
Hence if $f$ is even, then $g$ is odd and vice versa.
I think you'll have a hard time proving that the families you have are everything, unless you have some sort of differentiability assumption.
A: 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
A: You can take $g=1,$ then your equation reduces to $f(x+y)=f(x)+f(y).$ Without any additional assumption on $f,$ this functional equation has plenty of discontinuous solutions. See here 
Of course, once you add some regularity condition (such as continuity, differentiability, boundedness), you can get much more using what Vadim wrote in his post.
As to your question 2 - the same arguments apply, since this is a consequence of the general equation.
Since I do not know which condition you prefer to add, let me just mention that if $g$ is continuous, then equation $g(x+y)=2g(x)g(y)$ has solution $g(x)=0$ or $g(x)=2^{ax-1}.$ To see that, note that if $g(x_0)=0$ at some point then $g(x+x_0)=0$ everywhere. Alternatively, $g(2x)=2g^2(x)>0$ for all $x$ implies that the function $h(x)=\ln g(x)-1$ satisfies $h(x+y)=h(x)+h(y)$ and $h$ is continuos. So it is linear. Note, that I assumed that $f$ maps $\mathbb{C}$ into $\mathbb{R}$ but the situation does not change much if you switch to functions with complex range.
