Example 1. Try to calculate the sum of the distances from point $\mathrm{P}$ on the circumference of the circumscribed circle of a regular n-sided polygon to each vertex, and find its maximum and minimum values.
Solve as shown in Figure 2.
For a regular n-gon inscribed in a circle,
the radius of the circle is $R$, and $P$ is
a point on the arc $\overparen{A_{1} A_{n}}$, and
let $\angle \mathrm{POA}_{\mathrm{n}}=\alpha$,
by the Law of Sines, we have
$$
\begin{array}{l}
\mathrm{PA}_{1}=2 \mathrm{R} \cdot \sin \frac{\theta-\alpha}{2}, \\
\mathrm{PA}_{2}=2 \mathrm{R} \cdot \sin \frac{2 \theta-\alpha}{2},
\end{array}
$$
$$
P A_{n}=2 R \cdot \sin \frac{n \theta-\alpha}{2} \text {. }
$$

Thus, $\mathrm{s}=\sum_{k=1}^{n} P A_{k}=2 R \sum_{k=1}^{n} \sin \frac{k \theta-\alpha}{2}$.
$$
\begin{array}{l}
\text { According to formula (1), } \sum_{k=1}^{n} \sin \frac{k \theta-\alpha}{2} \\
=\frac{\sin \frac{(n+1) \theta-2 \alpha}{4} \cdot \sin \frac{n \theta}{4}}{\sin \frac{\theta}{4}} .
\end{array}
$$

Substituting $\theta=\frac{2 \pi}{n}$ into the above equation, we get
$$
\begin{aligned}
\mathrm{s}= & \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{PA}_{\mathrm{k}}=2 \mathrm{R} \sum_{\mathrm{k}=1}^{\mathrm{n}} \sin \frac{\mathrm{k} \theta-\alpha}{2} \\
& =2 \mathrm{R} \cdot \frac{\cos \left(\frac{\pi}{2 \mathrm{n}}-\frac{\alpha}{2}\right)}{\sin \frac{\pi}{2 \mathrm{n}}} . \\
& \because \mathrm{P} \text { is on } \mathrm{A}_{1} \mathrm{~A}_{\mathrm{n}}, \text { hence } \alpha \in\left(0, \frac{2}{\mathrm{n}}\right) .
\end{aligned}
$$

Clearly, when $\alpha=\frac{\pi}{n}$, i.e., $\mathrm{P}$ is at the midpoint of $\overparen{A_{1} A_{n}}$, $\mathrm{s}$ reaches its maximum value,
$$
\mathrm{s}_{\mathrm{max}}=\frac{2 R \cdot \cos 0}{\sin \frac{\pi}{2 n}}=2 \Omega \cdot \csc \frac{\pi}{2 n} \cdot
$$

When $\alpha=0$ or $\alpha=\frac{2 \pi}{n}$, i.e., $\mathrm{P}$ is at $\mathrm{A}_{1}$ or $\mathrm{A}_{0}$, $\mathrm{s}$ reaches its minimum value,
$$
s_{m i n}=2 R \cdot \operatorname{ctg}-\frac{\pi}{2 n} .
$$