13.4.21 ** Let $M$ be any point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, $F_{1}$ and $F_{2}$ be its two foci, and let $\angle F_{1} M F_{2}=\theta$.
(1) Find the maximum value of $\theta$; (2) Find the maximum value of $\sin \theta$.
(1) Applying the cosine rule in $\triangle M F_{1} F_{2}$:
$$
\begin{aligned}
\cos \theta & =\frac{\left|M F_{1}\right|^{2}+\left|M F_{2}\right|^{2}-\left|F_{1} F_{2}\right|^{2}}{2\left|M F_{1}\right| \cdot\left|M F_{2}\right|}=\frac{\left(\left|M F_{1}\right|+\left|M F_{2}\right|\right)^{2}-\left|F_{1} F_{2}\right|^{2}}{2\left|M F_{1}\right|\left|M F_{2}\right|}-1 \\
& =\frac{4 a^{2}-4 c^{2}}{2\left|M F_{1}\right| \cdot\left|M F_{2}\right|}-1 \quad \text { (where } c-\sqrt{a^{2}-b^{2}} \text { ) } \\
& =\frac{2 b^{2}}{\left|M F_{1}\right| \cdot\left|M F_{2}\right|}-1 \geqslant \frac{2 b^{2}}{\left(\frac{\left|M F_{1}\right|+\left|M F_{2}\right|}{2}\right)^{2}}-1=\frac{2 b^{2}}{a^{2}}-1 .
\end{aligned}
$$

When and only when $\left|M F_{1}\right|=\left|M F_{2}\right|$, i.e., $M$ is at the endpoint of the minor axis, the equality holds. By the monotonicity of the cosine function,
$$
\theta_{\max }=\arccos \left(\frac{2 b^{2}}{a^{2}}-1\right) \text {. }
$$
(2) Analysis 1: From (1), we know $\theta_{\max }=$ $\arccos \left(\frac{2 b^{2}}{a^{2}}-1\right)$. Therefore, when $\frac{2 b^{2}}{a^{2}}-1 \leqslant 0$, i.e., $a \geqslant \sqrt{2} b$, $\theta_{\text {max }} \geqslant \frac{\pi}{2}$, then taking $\theta=\frac{\pi}{2}$, $(\sin \theta)_{\text {max }}=1$; when $\frac{2 b^{2}}{a^{2}}-1>0$, i.e., $a<\sqrt{2} b$, $\theta_{\text {max }} \leqslant \frac{\pi}{2}$, then taking $\theta=\theta_{\text {max }}=\arccos \left(\frac{2 b^{2}}{a^{2}}-1\right)$,
$$
(\sin \theta)_{\max }=\sqrt{1-\left(\frac{2 b^{2}}{a^{2}}-1\right)^{2}}=\frac{2 b c}{a^{2}} \quad \text { (where } c=\sqrt{a^{2}-b^{2}} \text { ). }
$$

Analysis 2: We can set $\angle M F_{1} F_{2}=\alpha, \angle M F_{2} F_{1}=\beta$. By the sine rule, we get
$$
\frac{\left|F_{1} F_{2}\right|}{\sin \theta}=\frac{\left|M F_{2}\right|}{\sin \alpha}=\frac{\left|M F_{1}\right|}{\sin \beta}=\frac{\left|M F_{1}\right|+\left|M F_{2}\right|}{\sin \alpha+\sin \beta}
$$

i.e., $\frac{2 c}{\sin \theta}=\frac{2 a}{\sin \alpha+\sin \beta}$, where $c=\sqrt{a^{2}-b^{2}}$. Therefore, $\sin \theta=\frac{c}{a}(\sin \alpha+\sin \beta)=$ $\frac{2 c}{a} \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=\frac{2 c}{a} \cos \frac{\theta}{2} \cos \frac{\alpha-\beta}{2}$, thus $\sin \frac{\theta}{2}=\frac{c}{a} \cos \frac{\alpha-\beta}{2} \in\left[0, \frac{c}{a}\right]$. Therefore, $\sin ^{2} \theta=4 \sin ^{2} \frac{\theta}{2} \cos ^{2} \frac{\theta}{2}=4 \sin ^{2} \frac{\theta}{2}\left(1-\sin ^{2} \frac{\theta}{2}\right)=-4\left(\sin ^{2} \frac{\theta}{2}-\frac{1}{2}\right)^{2}+1$, hence when $\frac{c}{a} \geqslant \frac{\sqrt{2}}{2}$, i.e., $a \geqslant \sqrt{2} b$, taking $\sin \frac{\theta}{2}=\frac{\sqrt{2}}{2} ，(\sin \theta)_{\max }=1$; when $\frac{c}{a}<\frac{\sqrt{2}}{2}$, i.e., $a<\sqrt{2} b$, taking $\sin \frac{\theta}{2}=\frac{c}{a},(\sin \theta)_{\max }=2 \cdot \frac{c}{a} \sqrt{1-\left(\frac{c}{a}\right)^{2}}=\frac{2 b c}{a^{2}}$.