6-8 Given $\sin \alpha+\sin \beta=p, \cos \alpha+\cos \beta=q$, try to find the values of $\sin (\alpha+\beta)$ and $\cos (\alpha+\beta)$. (Beijing Mathematical Competition, 1963)
【Solution】First, from the known conditions, we know that the absolute values of $p, q$ cannot be greater than 2. Given:
$$
\begin{array}{l}
\sin \alpha+\sin \beta=p, \\
\cos \alpha+\cos \beta=q .
\end{array}
$$

Squaring both sides of (1) and (2) respectively, we get
$$
\begin{array}{l}
\sin ^{2} \alpha+\sin ^{2} \beta+2 \sin \alpha \sin \beta=p^{2}, \\
\cos ^{2} \alpha+\cos ^{2} \beta+2 \cos \alpha \cos \beta=q^{2} .
\end{array}
$$

Adding both sides of (3) and (4) respectively, we get
$$
2+2(\cos \alpha \cos \beta+\sin \alpha \sin \beta)=q^{2}+p^{2},
$$

which is $2[1+\cos (\alpha-\beta)]=q^{2}+p^{2}$.
Subtracting (3) from (4), we get
$$
\cos 2 \alpha+\cos 2 \beta+2 \cos (\alpha+\beta)=q^{2}-p^{2},
$$

which is $2 \cos (\alpha+\beta)[\cos (\alpha-\beta)+1]=q^{2}-p^{2}$.
Multiplying (1) by (2), we get
$$
\sin (\alpha+\beta)+\frac{1}{2}(\sin 2 \alpha+\sin 2 \beta)=p q,
$$

which is $\sin (\alpha+\beta)[1+\cos (\alpha-\beta)]=p q$.
Substituting (5) into (6) and (7) respectively, we get
$$
\begin{array}{l}
\left(p^{2}+q^{2}\right) \cos (\alpha+\beta)=q^{2}-p^{2}, \\
\left(p^{2}+q^{2}\right) \sin (\alpha+\beta)=2 p q .
\end{array}
$$

When $p, q$ are not both zero, we get
$$
\begin{array}{l}
\sin (\alpha+\beta)=\frac{2 p q}{p^{2}+q^{2}} ; \\
\cos (\alpha+\beta)=\frac{q^{2}-p^{2}}{q^{2}+p^{2}} .
\end{array}
$$

When $p=q=0$, from (1) we get $\sin \beta=-\sin \alpha$,
from (2) we get $\cos \beta=-\cos \alpha$.

Therefore,
$$
\begin{aligned}
\sin (\alpha+\beta) & =\sin \alpha \cos \beta+\cos \alpha \sin \beta \\
& =\sin \alpha(-\cos \alpha)+\cos \alpha(-\sin \alpha) \\
& =-\sin 2 \alpha, \\
\cos (\alpha+\beta) & =\cos \alpha \cos \beta-\sin \alpha \sin \beta \\
& =\cos \alpha(-\cos \alpha)-\sin \alpha(-\sin \alpha) \\
& =-\cos 2 \alpha .
\end{aligned}
$$

In this special case, the solution is indeterminate.