6-159 Functions $f, g: R \rightarrow R$ are not constant and satisfy
$$
\begin{array}{l}
f(x+y) \equiv f(x) g(y)+g(x) f(y), \\
g(x+y) \equiv g(x) g(y)-f(x) f(y),
\end{array} \quad x, y \in R .
$$

Find all possible values of $f(0)$ and $g(0)$.
[Solution] In the two identities, let $x=y=0$, we get
$$
f(0)=2 f(0) g(0) \text { and } g(0)=(g(0))^{2}-(f(0))^{2} \text {. }
$$

Since $\quad g(0) \neq \frac{1}{2}$,
otherwise from the second equation we get
$$
(f(0))^{2}=\frac{1}{4}-\frac{1}{2}<0 .
$$

Therefore, from the first equation, we get $f(0)=0$, and from the second equation, we get $g(0)=1$ or $g(0)=0$. However, the latter case is impossible, otherwise taking $y=0$ in the first identity, we have
$$
f(x) \equiv f(x) g(0)+g(x) f(0) \equiv 0,
$$

which contradicts the assumption that $f(x)$ is not a constant. Therefore, the only possibility is $f(0)=0, g(0)=1$. The functions $f(x)=\sin x, g(x)=\cos x$ are an example.