If $f$ and $g$ satisfy the sine/cosine addition formulae, then what is $g'(0)$?

1. From the first equation, putting $x=y=0$, we get $f(0)=0$.

2. From the second equation, putting $x=y=0$, we get $g(0)=g^2(0)$. So, either $g(0)=1$ or $g(0)=0$.

3. If $g(0)=0$ then from the first equation, putting $y=0$, we get $f(x)=0$, for all $x$, and then from the second equation, putting $y=0$, we get $g(x)=0$ for all $x$. From where you can compute that the derivative equals to zero.

Let us assume for the rest that $g(0)=1$. From the first equation, putting $x=0$, we get $f(-y)=-f(y)$. And from the second, putting $x=0$, we get $g(-y)=g(y)$.

So the equations are equivalent to

$$\begin{align}f(x+y)&=f(x)g(y)+f(y)g(x)\\g(x+y)&=g(x)g(y)-f(x)f(y)\end{align}$$

Since $g$ is even it is enough to compute the derivative from the right.

$$\begin{align}\lim_{y\rightarrow0^+}\frac{g(0+y)-g(0)}{y}&=\lim_{y\rightarrow0^+}\frac{g(0+y)-g(0)}{y}\\&=\lim_{y\rightarrow0^+}\frac{-2f(y/2)f(y/2)}{y}\\&=-\lim_{y\rightarrow0^+}\frac{f^2(y/2)}{y/2}\\&=-f(0^+)f'_{+}(0)\\&=0\end{align}$$

In the second equality we used the formula:

$$\begin{align}g(x)-g(y)&=-2f(\tfrac{x+y}{2})\,f(\tfrac{x-y}{2})\end{align}$$

To deduce it we use:

$$g(x)=g(\tfrac{x+y}{2}+\tfrac{x-y}{2})=g(\tfrac{x+y}{2})g(\tfrac{x-y}{2})-f(\tfrac{x+y}{2})f(\tfrac{x-y}{2})$$

$$g(y)=g(\tfrac{x+y}{2}-\tfrac{x-y}{2})=g(\tfrac{x+y}{2})g(\tfrac{x-y}{2})+f(\tfrac{x+y}{2})f(\tfrac{x-y}{2})$$

Subtracting the two equations we get

$$g(x)-g(y)=-2f(\tfrac{x+y}{2})f(\tfrac{x-y}{2})$$