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

Here is an outline of a proof I found as this question was recently reposted here on Math StackExchange.

Let $P(x,y)$ be the assertion of the first functional equation and $Q(x,y)$ be the assertion of the second functional equation.

Then $P(0,0)$ implies $f(0)=0$ and $Q(0,0)$ implies $g(0)=0$ or $g(0)=1$. By $Q(0,0)$ we have $$g(0)=f^2(x)+g^2(x)$$ for all $x\in\mathbb R$. So if $g(0)=0$ then we have $f(x)=g(x)=0$ for all $x$ and in particular $g'(0)=0$. Assume now that $g(1)=1$.

By $P(0,-y)$ we get $f(-y)=-f(y)$ for all $y$ and similarly $Q(0,y)$ implies $g(-y)=g(y)$ for all $y$. So $g$ is even and $f$ is odd. In particular, since the right-hand side limit of $f$ exists at $x=0$, we know that $f$ is also differentiable at $0$. In particular, $f$ is continuous at $0$.

Because of $f^2(x)+g^2(x)=1$, there exists a function $\theta:\mathbb R\to[-\pi,\pi[$ such that $$\big(f(x),g(x)\big)=\big(\sin(\theta(x)),\cos(\theta(x))\big)$$ for all $x\in\mathbb R$.

Now I need this

Lemma. We have $g>0$ in some neighborhood of $0$.
Proof. Because of $$1=f^2(x)+g^2(x)$$ and $f(0)=0$ and $f$ being continuous at $0$, we know that $g^2(x)\geq\frac14$ so $|g(x)|\geq\frac12$ in a neighborhood of $0$. Also, we have $$g(x/2)=g(x-x/2)=g(x)g(x/2)+f(x)f(x/2).$$ Now, in a neighborhood of $0$, we have $g(x)\neq0$ so that $$1=g(x)+\frac{f(x)f(x/2)}{g(x/2)}$$ which means that $$g(x)=1-\frac{f(x)f(x/2)}{g(x/2)}\geq 1-\frac{|f(x)f(x/2)|}{|g(x/2)|}\geq1-2 |f(x)f(x/2)|.$$ Again by continuity of $f$ at $0$ and $f(0)=0$, it follows that $g(x)>0$ at a neighborhood of $0$. $\square$

By the Lemma, we have $\theta(x)\in\left]-\frac\pi2,\frac\pi2\right[$ in a neighborhood of $0$. All the following statements are for $x$ in a neighborhood of $0$:

We have $\theta=\arcsin\circ f$. So, because $f(0)=0$ and $f$ is differentiable at $0$, we have $\theta(0)=0$ and $\theta$ is also differentiable at $0$. Hence we have by the chain rule $$g'(0)=\left.\frac{\mathrm d}{\mathrm dx}\right|_{x=0} \cos(\theta(x))=-\sin(\theta(0))\cdot\theta'(0)=-\sin(0)\cdot\theta'(0)=0\cdot\theta'(0)=0. $$

Answer 3

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.