Example 13 (Olympiad Training Problem from "Intermediate Mathematics" Issue 4, 2004) Determine all real numbers $\theta$ such that the sequence of complex numbers $\{\cos n \theta+\mathrm{i} \sin n \theta\}$ forms an arithmetic sequence.

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Let $a_{n}=\cos n \theta+\operatorname{isin} n \theta$, then $\left\{a_{n}\right\}$ is an arithmetic sequence, and $a_{n}-a_{n-1}$ is a constant value independent of $n$.
Also,
$$
\begin{aligned}
a_{n}-a_{n-1} & =[\cos n \theta-\cos (n-1) \theta]+\mathrm{i}[\sin n \theta-\sin (n-1) \theta] \\
& =2 \sin \frac{\theta}{2}\left(-\sin \frac{2 n-1}{2} \theta+\mathrm{i} \cos \frac{2 n-1}{2} \theta\right),
\end{aligned}
$$

Clearly, for $f(\theta)=-\sin \frac{2 n-1}{2} \theta+\mathrm{i} \sin \frac{2 n-1}{2} \theta$ to be independent of $n$, taking $n=1$ and $n=2$, we get $\sin \frac{\theta}{2}=\sin \frac{3 \theta}{2}, \cos \frac{\theta}{2}=\cos \frac{3 \theta}{2}$.
Thus, $\frac{3 \theta}{2}=\frac{\theta}{2}=2 k \pi$, i.e., $\theta=2 k \pi$.
It is easy to verify that when $\theta=2 k \pi$, $a_{n}=1$.
Therefore, the $\theta$ that makes $a_{n}$ an arithmetic sequence is $\theta=2 k \pi, k \in \mathbf{Z}$.