''Differential equation'' with known solution $\sin$ and $\cos$ I am given the following two two equations

$f,g : \mathbb{R} \to \mathbb{R}$ are differentiable on $\mathbb{R}$ and they satisfy $\forall x,y \in \mathbb{R}$ $$f(x+y) = f(x)g(y)+f(y)g(x)\\g(x+y)=g(x)g(y)-f(x)f(y)  $$ with $f(-x)=-f(x), \forall x \in \mathbb{R}$ and $f'(0)=1$.

I want to show that $f= \sin$ and $g = \cos$ 
My approach: My general idea was to show that $$f'=g, \  g'=-f$$ and $$f(0)=0, \ g(0)=1$$ 
If I'd manage to do that, then the solution follows because the above differential equation has unique solution $f= \sin, g = \cos$.

Since $f$ is an 'uneven' function it follows that $f(0)=-f(0) \implies f(0)=0$ so that's one check out of the above list $\checkmark$ 
Differentiating the first given expression I'd obtain $$f'(x+y)=f'(x)g(y)+f(y)g'(x) \implies f'(0)=f'(0)g(0)+f(0)g'(0)=g(0)=1 $$ 
So that takes care of $g(0)=1 \ \checkmark$
Now to the tricky part. I will differentiate both equations and obtain: $$f'(x+y)=f'(x)g(y) + f(y) g'(x) \\ g'(x+y)=g'(x)g(y)-f'(x)f(y) $$
It seems like it is crucial that $g'(0)=0$ in order to show the remaining two properties out of my list above. Which so far I didn't manage to do because all my steps are circular: $$f'(y)= f'(0)g(y)+f(y)g'(0)=g(y)+f(y)g'(0) \tag{*} $$
which would only help me if $g'(0)=0$, similarly $$g'(y)=g'(0)g(y)-f'(0)f(y)=g'(0)g(y)-f(y) $$
Which doesn't help me either, maybe there is a substitution I have to perform but I yet fail to see it.

Well, you can note that $$0=f(x-x)=f(x)g(-x)+f(-x)g(x)=f(x)g(-x)-f(x)g(x)=f(x)\bigl[g(x)-g(-x)\bigr].$$ This holds for all $x\in\Bbb R,$ and since $f'(0)\ne 0,$ then $f$ isn't the constant zero function. It follows that $g(x)=g(-x)$ whenever $f(x)\ne 0.$ In particular $f$ is non-zero in some punctured neighborhood about $x=0$ (why?), and clearly $g(0)=g(-0),$ so for all $x$ sufficiently close to $0,$ we have $g(x)=g(-x).$ From this, you can show that $g'(0)=0$ directly, using a symmetric derivative.
