TITLE: functional equations for trigonometric functions: $f(x+y)=f(x)f(y)-g(x)g(y)$ and $g(x+y)=f(x)g(y)+g(x)f(y)$
QUESTION [7 upvotes]: It is well known that the following system of functional equations:
$\begin{cases}
f(x+y)=f(x)f(y)-g(x)g(y) \\
g(x+y)=f(x)g(y)+g(x)f(y)
\end{cases}$
admit the solution $(f,g)=(\cos,\sin)$. Are there any other solutions (besides the trivial $(0,0)$)? i.e. do these identities characterize the trigonometric functions? If not, what additional identities are required?

REPLY [2 votes]: If you let $h(x)=f(x)+ig(x)$, your system reduces to $h(x+y)=h(x)h(y)$. It's not too hard to check that the continuous solutions to this take the form $h(t)=e^{\alpha t}$ for some $\alpha \in \Bbb{C}$, so the continuous solutions to the original system are of the form $f(t)=e^{bt}\cos at$, $g(t)=e^{bt}\cos at$ for some real constants $a,b$.
So Qiaochu's derivative condition is sufficient, if you're careful (i.e., if you were just ruling out $(f,g)=(\cos at, \sin at)$ you might think it was enough to specify that $g'(0)=1$, but in reality you also need to specify that $f'(0)=0$).