Example 6 (5th Hope Cup - Senior High School Grade 2 Question) In the sequence $\left\{a_{n}\right\}$, $a_{1}=a(0<a<1), a_{n+1}=$ $\sqrt{\frac{1-\sqrt{1-a_{n}^{2}}}{2}}$, find the general term formula of $\left\{a_{n}\right\}$.
Analysis: Noticing $\sqrt{1-a_{n}^{2}}$ in the problem, it naturally leads to the idea of using a trigonometric substitution.
Solution: Given $0<a<1$, we can set $a_{1}=a=\sin \theta\left(0<\theta<\frac{\pi}{2}\right.$ and $\left.\theta=\arcsin a\right)$, then
$$
\begin{array}{l}
a_{2}=\sqrt{\frac{1-\sqrt{1-\sin ^{2} \theta}}{2}}=\sqrt{\frac{1-\cos \theta}{2}}=\sin \frac{\theta}{2}, a_{3}=\sqrt{\frac{1-\sqrt{1-\sin ^{2} \frac{\theta}{2}}}{2}}= \\
\sqrt{\frac{1-\cos \frac{\theta}{2}}{2}}=\sin \frac{\theta}{4} .
\end{array}
$$

By analogy, $a_{n}=\sin \frac{\theta}{2^{n-1}}$, which can be proven by mathematical induction. (The following is omitted)
Evaluation: There is a class of problems where the given recurrence relation aligns with a certain trigonometric formula or trigonometric identity, and the range of the corresponding terms also matches a certain trigonometric function (for example, $\sin \alpha$ corresponds to the range $[-1,1]$, and $\sin ^{2} \alpha$ corresponds to the range $[0,1]$). In such cases, experienced students will decisively use a trigonometric substitution. As in this problem, the given recurrence relation is quite complex, and solving it without using a trigonometric substitution would be very cumbersome.