6.8 Given $\sin \alpha+\sin \beta=p, \cos \alpha+\cos \beta=q$, try to find the values of $\sin (\alpha+\beta)$ and $\cos (\alpha+\beta)$.

(China Beijing Mathematical Competition, 1963)
[Solution] First, from the given 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 (3) and (4) respectively, we get
$$2+2(\cos \alpha \cos \beta+\sin \alpha \sin \beta)=q^{2}+p^{2}$$

That is,
$$2[1+\cos (\alpha-\beta)]=q^{2}+p^{2}$$

Subtracting (3) from (4), we get
$$\cos 2 \alpha+\cos 2 \beta+2 \cos (\alpha+\beta)=q^{2}-p^{2}$$

That is,
$$2 \cos (\alpha+\beta)[\cos (\alpha-\beta)+1]=q^{2}-p^{2}$$

Multiplying (1) and (2), we get
$$\sin (\alpha+\beta)+\frac{1}{2}(\sin 2 \alpha+\sin 2 \beta)=p q$$

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

In this special case, the solution is indeterminate.