Wait for the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the maximum value of the area of an inscribed triangle.
Let $\triangle A B C$ be an inscribed triangle of the ellipse, with the coordinates of the three vertices being $A(a \cos \alpha, b \sin \alpha), B(a \cos \beta, b \sin \beta), C(a \cos \gamma, b \sin \gamma)(0 \leqslant \gamma<\beta<\alpha<2 \pi)$. The equation of $l_{A B}$ is $b(\sin \beta-\sin \alpha) x-a(\cos \beta-\cos \alpha) y+a b \sin (\alpha-\beta)=0$. Therefore,
$$
\begin{aligned}
S= & \frac{1}{2} \sqrt{a^{2}(\cos \alpha-\cos \beta)^{2}+b^{2}(\sin \alpha-\sin \beta)^{2}} \\
& \cdot \frac{|b(\sin \beta-\sin \alpha) a \cos \gamma-a(\cos \beta-\cos \alpha) b \sin \gamma+a b \sin (\alpha-\beta)|}{\sqrt{b^{2}(\sin \beta-\sin \alpha)^{2}+a^{2}(\cos \beta-\cos \alpha)^{2}}} \\
= & \frac{1}{2} a b|(\sin \beta-\sin \alpha) \cos \gamma-(\cos \beta-\cos \alpha) \sin \gamma+\sin (\alpha-\beta)| \\
= & \frac{1}{2} a b|\sin (\beta-\gamma)+\sin (\gamma-\alpha)+\sin (\alpha-\beta)|,
\end{aligned}
$$

Let $\theta_{1}=\alpha-\beta, \theta_{2}=\beta-\gamma$, then $\alpha-\gamma=\theta_{1}+\theta_{2}$, thus
$$
\begin{aligned}
S & =\frac{1}{2} a b\left|\sin \theta_{1}+\sin \theta_{2}-\sin \left(\theta_{1}+\theta_{2}\right)\right| \\
& =\frac{1}{2} a b\left|2 \sin \frac{\theta_{1}+\theta_{2}}{2} \cos \frac{\theta_{1}-\theta_{2}}{2}-2 \sin \frac{\theta_{1}+\theta_{2}}{2} \cos \frac{\theta_{1}+\theta_{2}}{2}\right| \\
& =2 a b\left|\sin \frac{\theta_{1}+\theta_{2}}{2} \sin \frac{\theta_{1}}{2} \sin \frac{\theta_{2}}{2}\right| \\
& \leqslant 2 a b\left(\frac{\sin ^{2} \frac{\theta_{1}+\theta_{2}}{2}+\sin ^{2} \frac{\theta_{1}}{2}+\sin ^{2} \frac{\theta_{2}}{2}}{3}\right)^{\frac{3}{2}}+\frac{2}{2 \sqrt{3}} a b\left(\frac{1-\cos \left(\theta_{1}+\theta_{2}\right)}{2}+\frac{1-\cos \theta_{1}}{2}+\frac{1-\cos \theta_{2}}{2}\right)^{\frac{3}{2}} \\
& =\frac{1}{3 \sqrt{6}} a b\left(3-\cos \theta_{1}-\cos \theta_{2}-\cos \left(\theta_{1}+\theta_{2}\right)\right)^{\frac{3}{2}} \\
& =\frac{1}{3 \sqrt{6}} a b\left(3-2 \cos \frac{\theta_{1}+\theta_{2}}{2} \cos \frac{\theta_{1}-\theta_{2}}{2}-2 \cos ^{2} \frac{\theta_{1}+\theta_{2}}{2}+1\right)^{\frac{3}{2}} \\
& =\frac{1}{3 \sqrt{6}} a b\left[\frac{9}{2}-2\left(\cos \frac{\theta_{1}+\theta_{2}}{2}+\frac{1}{2} \cos \frac{\theta_{1}-\theta_{2}}{2}\right)^{2}-\frac{1}{2} \sin ^{2} \frac{\theta_{1}-\theta_{2}}{2}\right]^{\frac{3}{2}} \\
& \leqslant \frac{1}{3 \sqrt{6}} a b\left(\frac{9}{2}\right)^{\frac{3}{2}}=\frac{3 \sqrt{3}}{4} a b .
\end{aligned}
$$

The maximum value is achieved when $\cos \theta_{1}=\cos \theta_{2}=\cos \left(\theta_{1}+\theta_{2}\right)$, and $\cos \frac{\theta_{1}+\theta_{2}}{2}+\frac{1}{2} \cos \frac{\theta_{1}-\theta_{2}}{2}=\sin \frac{\theta_{1}-\theta_{2}}{2}=0$, i.e., $\theta_{1}=\theta_{2}=\frac{2 \pi}{3}$. At this time, $\alpha=\beta+\frac{2 \pi}{3}, \beta=\gamma+\frac{2 \pi}{3}$.