In the last article, we explored what a single functional equation could tell us about a function — enough, in one case, to pin the function down almost completely. Here's another functional equation to explore, and this one turns out to characterize the sine function.
Suppose we know the following about a function \(f\):
$$\begin{align*} f(x+y)&=f(x)f(a-y)+f(y)f(a-x)\\[12pt] \text{and }f(a)&=1\\[24pt] f(x)&=f(x+0)=f(x)f(a)+f(0)f(a-x)\\[12pt] &=f(x)+f(0)f(a-x)\\[12pt] \implies f(0)&=0\\[24pt] f(a+x)&=f(a)f(a-x)+f(x)f(0)\\[12pt] &=f(a-x)\\[24pt] f(0)&=f(x-x)=f(x)f(a-x)+f(-x)f(a+x)\\[12pt] &=f(a+x)\left[f(x)+f(-x)\right]=0\\[12pt] \implies f(-x)&=-f(x)\\[24pt] f(2a)&=f(a)f(0)+f(a)f(0)=0\\[12pt] f(3a)&=f(2a+a)\\[12pt] &=f(2a)f(0)+f(a)f(-a)=-1\\[12pt] f(4a)&=2f(2a)f(-a)=0\\[24pt] f(x+4a)&=f(x)f(-3a)+f(4a)f(a-x)\\[12pt] &=f(x)\\[12pt] \implies f &\text{ periodic with }4a\text{ as the period where }\\[6pt] a&\text{ is the smallest positive solution of }f(a)=1 \end{align*}$$Next, let's save ourselves some writing by putting \(g(x)=f(a-x)\). So far, we have
$$\begin{align*} f(x+y)&=f(x)g(y)+f(y)g(x)\\[12pt] f\text{ and }&g \text{ are both periodic with period }4a\\[12pt] f(-x)&=-f(x)\qquad g(-x)=g(x)\\[12pt] f(0)&=g(a)=0\\[12pt] g(0)&=f(a)=1 \end{align*}$$ $$\begin{align*} g(x+y)&=f(a-(x+y))\\[12pt] &=f(a+(x+y))=f((a+x)+y)\\[12pt] &=f(a+x)g(y)+g(a+x)f(y)\\[12pt] &=g(x)g(y)-f(x)f(y)\\[24pt] \implies 1&=g(0)=g(x-x)\\[12pt] &=g(x)g(-x)-f(x)f(-x)\\[12pt] &=g(x)^2+f(x)^2 \end{align*}$$It is, I think, truly remarkable that we can show just from this that \(f'(x)=\alpha g(x)\) and \(g'(x)=-\alpha f(x)\) for some value of \(\alpha\).
$$\begin{align*} f'(x)&=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\\[12pt] &=\lim_{h\to 0}\frac{f(x)g(h)+f(h)g(x)-f(x)}{h}\\[12pt] &=g(x)\lim_{h\to0}\frac{g(h)-1}{h}+f(x)\lim_{h\to0}\frac{f(h)}{h}\\[12pt] &=\alpha g(x)+\beta f(x)\\[24pt] g'(x)&=\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\\[12pt] &=\lim_{h\to 0}\frac{g(x)g(h)-f(h)f(x)-f(x)}{h}\\[12pt] &=-f(x)\lim_{h\to0}\frac{f(h)}{h}+g(x)\lim_{h\to0}\frac{g(h)-1}{h}\\[12pt] &=-\alpha f(x)+\beta g(x) \end{align*}$$We already know that \(f(x)^2\leqslant 1\) and \(g(x)^2\leqslant 1\), so that \(f\) must have a max at \(a\). This means that
$$\begin{align*} &f'(a)=0\\[12pt] \implies &\alpha g(a)+\beta f(a)=0\\ \implies &\beta=0 \end{align*}$$At this point, the functions \(f\) and \(g\) look just like sine and cosine. In fact
$$\begin{align*} f(x)=\sin\alpha x \end{align*}$$is a solution to the original functional equation. If we add in the requirement that \(f'(0)=1\), we will have \(\alpha=1\) and \(f(x)=\sin x\) as a solution, with \(2a=\pi\). Of course we know from our work on the sine function that
$$\begin{align*} \alpha=1\iff 2a=\pi \end{align*}$$but I don't believe we can deduce this from the functional equation alone.