Example 9 As shown in Figure 2.10.9, within a unit circle, the vertex angle of sector $A O B$ varies within $\left(0, \frac{\pi}{2}\right)$, and $P Q R S$ is an inscribed square in this sector. Try to find the minimum value of $O S$.

 untranslated part:
 
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Consider $O$ as the pole and the ray $O A$ as the polar axis to establish a polar coordinate system. Let $S(\rho, \theta), R(1, \alpha)\left(0<\theta, \alpha<\frac{\pi}{2}\right)$, then
$$
\begin{array}{l}
|P S|=\rho \cdot \sin \theta,|O P|=\rho \cdot \cos \theta, \\
|R O|=\sin \alpha,|O Q|=\cos \alpha
\end{array}
$$

Also,
$$
|P Q|=\cos \alpha-\rho \cdot \cos \theta,|P Q|=|S P|
$$

Thus,
$$
\cos \alpha-\rho \cdot \cos \theta=\rho \cdot \sin \theta
$$

And,
$$
\begin{array}{l} 
\cos \alpha=|O Q|=\sqrt{O R^{2}-R Q^{2}}=\sqrt{1-\rho^{2} \cdot \sin ^{2} \theta} \\
\sqrt{1-\rho^{2} \cdot \sin ^{2} \theta}=\rho(\sin \theta+\cos \theta)
\end{array}
$$

Therefore,
$$
\begin{array}{l}
\rho^{2}=\frac{1}{\sin ^{2} \theta+2 \sin \theta \cdot \cos \theta+1}=\frac{2}{\sqrt{5} \sin (2 \theta+\varphi)+3} \geqslant \\
\frac{2}{\sqrt{5}+3}=\left(\frac{\sqrt{5}-1}{2}\right)^{2} \\
\end{array}
$$

Hence $\rho \geqslant \frac{\sqrt{5}-1}{2}$, which means the minimum value of $O S$ is $\frac{\sqrt{5}-1}{2}$.
For example 8, the boundedness of trigonometric functions can also be used.
Let $D E=x, \angle A E D=\alpha, \angle A D E=\beta$. From the given conditions, $\angle A C B=90^{\circ}, S_{\triangle A B C}=30$, $S_{\triangle A D E}=\frac{1}{2} A D \cdot A E \cdot \sin A=15$, so $A D \cdot A E \cdot \sin A=30$. By the Law of Sines,
$$
x: \sin A=A D: \sin \alpha=A E: \sin \beta
$$

Thus,
$$
x=\frac{A D \cdot \sin A}{\sin \alpha}, x=\frac{A E \cdot \sin A}{\sin \beta}
$$

Therefore,
$$
x^{2}=\frac{A D \cdot A E \cdot \sin ^{2} A}{\sin \alpha \cdot \sin \beta}=\frac{60 \sin A}{\cos (\alpha-\beta)+\cos A}
$$

So when $\cos (\alpha-\beta)=1$, $x^{2}$ reaches its minimum value of 12, which means the minimum length of $D E$ is $2 \sqrt{3}$.