If $f(x+y)=f(x)f(y)-g(x)g(y)$ and $g(x+y)=f(x)g(y)+f(y)g(x)$, with $f'(0)=0$, determine $[f(x)]^2+[g(x)]^2$ Given the expressions:
$f(x+y)=f(x)f(y)-g(x)g(y)$
$g(x+y)=f(x)g(y)+f(y)g(x)$
the exercise is to show that $[f(x)]^2+[g(x)]^2$ is constant for all real $x$ and determine its value, knowing that $f$ and $g$ are real differentiable non-constant functions, and that $f'(0)=0$.
I realized it looks like the $\sin$ and $\cos$ functions, so the answer must be $1$. To prove something that way, I tried showing that $f$ and $g$ were always on the interval $[-1,1]$.
I have also tried to derivate each expression and plug in $x=y=0$ or only $y=0$, but was unable to develop the solution.

If we let $\phi(x)=f(x)+ig(x)$, then $\phi$ satisfies a functional equation 
$$
\phi(x+y)=\phi(x)\phi(y).
$$ This gives $\phi(x)=\phi(x)\phi(0)$, so either $\phi \equiv 0$ or $\phi(0)=1$. Since $\phi \equiv 0$ is a trivial solution, we assume $\phi(0)=1$. Differentiating with $y$ and plugging $y=0$, we obtain $\phi'(x)=\phi'(0)\phi(x)$. Now, define $u(x)=e^{-\phi'(0)x}\phi(x)$ and observe that
$$
u'(x)=e^{-\phi'(0)x}\left(\phi'(x)-\phi'(0)\phi(x)\right)=0.
$$ This gives $u(x) = u(0)=\phi(0)=1$ and hence $\phi(x) = e^{\phi'(0)x}$. Since $$\phi'(0)=f'(0)+ig'(0)=ig'(0)$$ it follows $\phi(x)  = e^{ig'(0)x}=e^{i\theta x}$ for some $\theta\in\Bbb R$, hence by Euler's identity $$
\left(f(x)\right)^2+\left(g(x)\right)^2 =\cos^2(\theta x)+\sin^2(\theta x) =1.
$$ (Also note that trivial solution $\phi =0$ gives $\left(f(x)\right)^2+\left(g(x)\right)^2 =0$.)
