Example 15 Find the sum: $\sum_{k=0}^{2 n} \cos \left(\alpha+\frac{2 k \pi}{2 n+1}\right)$.
Let $z_{k}=\cos \left(\alpha+\frac{2 k \pi}{2 n+1}\right)+i \sin \left(\alpha+\frac{2 k \pi}{2 n+1}\right) (k=0,1,2, \cdots, 2 n)$.
Take the $(2 n+1)$-th roots of unity $\varepsilon_{k}=\cos \frac{2 k \pi}{2 n+1}+i \sin \frac{2 k \pi}{2 n+1} (k=0,1,2, \cdots, 2 n)$,
then
$$
\begin{aligned}
W & =z_{0}+z_{1}+z_{2}+\cdots+z_{2 n}=z_{n}+z_{0} \varepsilon_{1}+z_{0} \varepsilon_{2}+\cdots+z_{0} \varepsilon_{2 n} \\
& =z_{0}\left(1+\varepsilon_{1}+\varepsilon_{2}+\cdots+\varepsilon_{2 n}\right)=0 .
\end{aligned}
$$

The sum $\sum_{k=0}^{2 n} \cos \left(\alpha+\frac{2 k \pi}{2 n+1}\right)$ is the real part of the complex number $W$, hence $\sum_{k=0}^{2 n} \cos \left(\alpha+\frac{2 k \pi}{2 n+1}\right)=0$.
In particular, when $\alpha=0, n=3$, we have $\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}=\frac{1}{2}$, which is an IMO-5 problem. From the above proof process, it can also be seen that $\sum_{k=0}^{2 n} \sin \left(\alpha+\frac{2 k \pi}{2 n+1}\right)=0$.