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

Question

The question is:

Question. Let $f,g:\mathbb R\to\mathbb R$ be two functions that satisfy$$f(x-y)=f(x)\cdot g(y)-f(y)\cdot g(x)$$ and $$g(x-y)=g(x)\cdot g(y)+f(x)\cdot f(y)$$ for all $x,y \in \mathbb{R} $.

If the right hand derivative at $x=0$ exists for $f(x)$, then what is $g'(0)$ ?

My try:

By some simple substitutions I figured out that $f(0)=0$ and $g(0)=1$. If in the second equation, we put $x=y$, it will give $g(0)=(g(x))^2+(f(x))^2$. If $g(0)=0$, sum of the two squares becomes $0$ which implies the squares themselves are zero, I neglected $g(x)=f(x)=0$ as a trivial solution and hence took $g(0)=1$. But how do I proceed after this?

Answer 1

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

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

Answer 2

Those look like the functional definitions of sin and cos respectively, see Trig functions.

$\sin(x - y) = \sin(x)\cos(y) - \cos(x)\sin(y)$

$\cos(x - y) = \cos(x)\cos(y) - \sin(x)\sin(y)$

You would need a constant $c $ to be able to meet the entire range of functions.

But, $g(0) = c \cos(0) = c = 1$

So you would simply have $f(x) = \sin(x)$ and $g(x) = \cos(x)$ and it would be simple to find $g'(x)$.

Let me know if this helps.