Example 5 Consider the double-angle sine formula $\sin 2 \alpha=2 \sin \alpha \cos \alpha$,
First, transform to get: $\cos \alpha=\frac{1}{2} \cdot \frac{\sin 2 \alpha}{\sin \alpha}$, and write it as a recursive split term:
$$
\cos 2^{k} \alpha=\frac{1}{2} \cdot \frac{\sin 2^{k+1} \alpha}{\sin 2^{k} \alpha}, k \in \mathbf{N} \text {. }
$$

This yields the equation:
$$
\cos \alpha \cos 2 \alpha \cos 4 \alpha \cdots \cos 2^{n} \alpha=\frac{1}{2^{n}} \cdot \frac{\sin 2^{n+1} \alpha}{\sin \alpha} .
$$