2021 CSMC contest question This was the last question on the Canadian senior mathematics contest and the contest ended 4 days ago and it said to wait at least 48 hours before discussing any questions online.
A pair of functions $f(x)$ and $g(x)$ is called a Payneful pair if
(i) $f(x)$ is a real number for all real numbers $x$,
(ii) $g(x)$ is a real number for all real numbers $x$,
(iii) $f(x + y) = f(x)g(y) + g(x)f(y)$ for all real numbers $x$ and $y$,
(iv) $g(x + y) = g(x)g(y) - f(x)f(y)$ for all real numbers $x$ and $y$, and
(v) $f(a) ≠ 0$ for some real number $a$.
For every Payneful pair of functions $f(x)$ and $g(x)$:
(a) Determine the values of $f(0)$ and $g(0)$.
(b) If $h(x) = (f(x))^2 + (g(x))^2$ for all real numbers $x$, determinate the value of $h(5)h(-5)$
(c) If $-10 ≤ f(x) ≤ 10$ and $-10 ≤ g(x) ≤ 10$ for all real numbers $x$, determine the value of $h(2021)$.
I noticed that the two functions seem very similar to the sine and cosine functions and got $f(0) = 0, g(0) = 1$ (I did not obtain $f(0) = 0, g(0) = 1)$ by assuming that f and g are sine and cosine functions but I noticed it would match to a sine and cosine function), but I have no idea how to solve parts b and c (Assuming my solution  for part a is correct)
 A: 
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.
A: 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 $x$.
