Example 4 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{n+1}=a_{n}^{3}-3 a_{n}$, and $a_{1}=\frac{3}{2}$. Find the general term formula of $\left\{a_{n}\right\}$.
Solve: From $a_{n+1}=a_{n}^{3}-3 a_{n} \Rightarrow \frac{a_{n+1}}{2}=4\left(\frac{a_{n}}{2}\right)^{3}-3 \times \frac{a_{n}}{2}$.
Let $b_{n}=\frac{a_{n}}{2}$. Then
$$
b_{n+1}=4 b_{n}^{3}-3 b_{n}, b_{1}=\frac{3}{4} \text {. }
$$

Using the triple angle formula $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$.
If $b_{n}=\cos \theta_{n}$, then
$$
\begin{array}{l}
b_{n+1}=4 b_{n}^{3}-3 b_{n}=4 \cos ^{3} \theta_{n}-3 \cos \theta_{n}=\cos 3 \theta_{n} \\
\Rightarrow \theta_{n+1}=3 \theta_{n} .
\end{array}
$$

Since $\theta_{1}=\arccos \frac{3}{4}$, we have
$$
\theta_{n}=3^{n-1} \theta_{1}=3^{n-1} \arccos \frac{3}{4} \text {. }
$$

Thus, $b_{n}=\cos \left(3^{n-1} \arccos \frac{3}{4}\right)$,
$$
a_{n}=2 b_{n}=2 \cos \left(3^{n-1} \arccos \frac{3}{4}\right) \text {. }
$$