Question 2 As shown in Figure 2, find the maximum value of the area of an inscribed isosceles trapezoid in a semicircle (the radius of the circle is 1).


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Analysis: From an algebraic perspective. Let
Figure 2 $O E=x(0<x<1)$, draw $C E \perp A B$, with the foot of the perpendicular being $E$, then the area of the isosceles trapezoid $A B C D$ is
$$S=\frac{1}{2}(|A B|+|C D|) \cdot|C E|,$$

which is $S=(1+x) \sqrt{1-x^{2}}$.
The extremum of this function can be found using the AM-GM inequality for four variables. In fact,
$$\begin{array}{l}
S^{2}=(1+x)^{2}\left(1-x^{2}\right) \\
=\frac{1}{3}(1+x)^{3}(3-3 x) \\
\leqslant \frac{1}{3} \cdot\left[\frac{(1+x)+(1+x)+(1+x)+(3-3 x)}{4}\right]^{4} \\
=\frac{27}{16}
\end{array}$$

Taking the square root, we get $S \leqslant \frac{3 \sqrt{3}}{4}$.
When $1+x=3-3 x$, i.e., $x=\frac{1}{2}$, $S_{\text {max }}=\frac{3 \sqrt{3}}{4}$.
Alternatively, we can consider from a trigonometric perspective. Let $\angle A O D = \theta$, then the area of the isosceles trapezoid $A B C D$ is
$$\begin{aligned}
S & =\frac{1}{2} \sin \theta+\frac{1}{2} \sin \theta+\frac{1}{2} \sin (\pi-2 \theta) \\
& =\sin \theta+\frac{1}{2} \sin 2 \theta
\end{aligned}$$

Transforming and using the AM-GM inequality for four variables, we get
$$\begin{aligned}
S & =(1+\cos \theta) \sin \theta \\
& =2 \cos ^{2} \frac{\theta}{2} \cdot 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}=4 \cos ^{3} \frac{\theta}{2} \cdot \sin \frac{\theta}{2} \\
& =\frac{4}{\sqrt{3}} \cdot \sqrt{\cos ^{2} \frac{\theta}{2} \cdot \cos ^{2} \frac{\theta}{2} \cdot \cos ^{2} \frac{\theta}{2} \cdot 3 \sin ^{2} \frac{\theta}{2}} \\
& \leqslant \frac{4}{\sqrt{3}} \cdot \sqrt{\left[\frac{\cos ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}+\cos ^{2} \frac{\theta}{2}+3 \sin ^{2} \frac{\theta}{2}}{4}\right]^{4}} \\
& =\frac{3 \sqrt{3}}{4}
\end{aligned}$$