Example 3
Let the side length of a regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$ be $a$, find its covering radius.
We prove that the circumcircle $\odot O$ of the regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$ is its minimum covering circle. The usual steps to prove that $\odot O$ is the minimum covering circle are: (1) $\odot O$ is a covering circle; (2) the radius of any covering circle $\geqslant$ the radius of $\odot O$. We will use this method to prove the problem.

Clearly, $\odot O$ covers this regular $n$-sided polygon. Therefore, the diameter of any circle covering the regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$ $\geqslant$ $A_{1} A_{m+1} = \odot O$'s diameter, which means $\odot O$ is the minimum covering circle.

When $n$ is an odd number $2 m-1$, it is easy to see that $\triangle A_{1} A_{m} A_{m+1}$ is an isosceles triangle with an acute angle: $\angle A_{m} A_{1} A_{m+1} = \frac{\pi}{2 m-1}$, so $\triangle A_{1} A_{m} A_{m+1}$ is an acute triangle. According to Example 2, its minimum covering circle is $\odot O$, thus the minimum covering circle of the regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$ is $\odot O$.

In summary, the circumcircle $\odot O$ of the regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$ is the minimum covering circle. Therefore, the covering radius is the radius $r$ of $\odot O$, and it is easy to know that
$$
r = \frac{a}{2 \sin \frac{\pi}{n}}.
$$