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?
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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 $f'(0)=0$).
$(f, g) = (\cos a(t), \sin a(t))$ also solves this system for any solution $a(t)$ to the Cauchy functional equation $a(x + y) = a(x) + a(y)$. In addition to the trivial solutions $a(t) = at$, there are pathological solutions that can be constructed using the axiom of choice. The pathological solutions can be ruled out by mild hypotheses, e.g. I think $f, g$ measurable suffices (certainly $f, g$ continuous suffices). To rule out $a(t) = at$ I think the easiest thing to do is to impose a condition on derivatives.