2021 CSMC contest question

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It would take me a lot of time to type it back in Latex so I have used pictures here, please apologize me for using pictures

Surely when I find time I will convert this into latex.

You cannot assume $f, g$ are sine and consine. For example, if $f$ and $g$ is a Payneful pair, so is $-f$ and $g$. Therefore they are not unique.

(a) Let $x=y=0$ in the identities, then $$f(0)=2f(0)g(0)\Leftrightarrow f(0)(1-2g(0))=0$$ $$g(0)=g(0)^2-f(0)^2$$ If $g(0)=\frac{1}{2}$, then $f(0)^2=g(0)^2-g(0)<0$. Thus $f(0)=0$, and $$g(0)=g(0)^2\Leftrightarrow g(0)=0 \text{ or }g(0)=1$$ If $g(0)=0$, then $f(a)=f(a)g(0)+g(a)f(0)=0$. Thus $g(0)=1$.

(b) It's easy to show $$h(x+y)=h(x)h(y)$$ Thus by part (a), $$h(5)h(-5)=h(0)=1$$

(c) The conditions imply that $h(x)$ is bounded. If $|h(x)|>1$, then $|h(nx)|=|h(x)^n|\rightarrow\infty$ as $n\rightarrow\infty$. Therefore $|h(x)|\le 1$. Note that $h(x)h(-x)=h(0)=1$, we have $|h(x)||h(-x)|=1$ and by $|h(x)|, |h(-x)|\le 1$, we must have $|h(x)|=|h(-x)|=1$ for arbitrary $x$. And since $h(x)\ge 0$ as a sum of squares, we must have $h(x)=1$ for all $x$.