20.5. (New York, 76). Functions $f, g: R \rightarrow R$ are non-constant and satisfy two identities

$$
\begin{aligned}
& f(x+y)=f(x) g(y)+g(x) f(y) \\
& g(x+y)=g(x) g(y)-f(x) f(y)
\end{aligned}
$$

$x, y \in \mathbf{R}$. Find all possible values of $f(0)$ and $g(0)$.
20.5. Substituting $x=y=0$ into each of the two identities satisfied by the non-constant functions $f(x), g(x)$, we obtain two equations

$$
f(0)=2 f(0) g(0) \text { and } g(0)=(g(0))^{2}-(f(0))^{2}
$$

Since $g(0) \neq 1 / 2$ (otherwise, from the second equation it would follow that $\left.(f(0))^{2}=1 / 4-1 / 2<0\right)$, from the first equation we have $f(0)=0$, and from the second we get that either $g(0)=1$, or $g(0)=0$. However, the latter case is impossible, because otherwise, by the first identity with $y=0$, the function

$$
f(x) \equiv f(x) g(0)+g(x) f(0) \equiv 0
$$

would be constant. Thus, the only possible values are $f(0)=0, g(0)=1$, which are realized, for example, by the functions

$$
f(x)=\sin x, g(x)=\cos x
$$