4-15 Given that $\operatorname{tg} \alpha$ and $\operatorname{tg} \beta$ are the roots of $x^{2}+p x+q=0$, try to find
$$
\sin ^{2}(\alpha+\beta)+p \sin (\alpha+\beta) \cos (\alpha+\beta)+q \cos ^{2}(\alpha+\beta) \text { }
$$

value.
[Solution] By the relationship between roots and coefficients, we have
$$
\operatorname{tg} \alpha + \operatorname{tg} \beta = -p \text{ and } \operatorname{tg} \alpha \operatorname{tg} \beta = q,
$$

Therefore, $\operatorname{tg}(\alpha+\beta) = \frac{-p}{1-q}$.
Thus, the value we seek is
$$
\begin{aligned}
& \sin ^{2}(\alpha+\beta) + p \sin (\alpha+\beta) \cos (\alpha+\beta) + q \cos ^{2}(\alpha+\beta) \\
= & \cos ^{2}(\alpha+\beta)\left[\operatorname{tg}^{2}(\alpha+\beta) + p \operatorname{tg}(\alpha+\beta) + q\right] \\
= & \frac{1}{1+\operatorname{tg}^{2}(\alpha+\beta)}\left[\operatorname{tg}^{2}(\alpha+\beta) + p \operatorname{tg}(\alpha+\beta) + q\right] \\
= & \frac{1}{1+\frac{p^{2}}{(1-q)^{2}}}\left[\frac{p^{2}}{(1-q)^{2}} - \frac{p^{2}}{1-q} + q\right] \\
= & q .
\end{aligned}
$$