9. Let $0<\theta<\pi$, find the maximum value of $y=\sin \frac{\theta}{2}(1+\cos \theta)$
9. Eliminate the trigonometric function through substitution.

Let $\sin \frac{\theta}{2}=x, x \in(0,1)$, then $y=2 \sin \frac{\theta}{2}\left(1-\sin ^{2} \frac{\theta}{2}\right)=2 x\left(1-x^{2}\right)$,
transforming we get $\quad 1=x^{2}+\frac{y}{2 x}=x^{2}+\frac{y}{4 x}+\frac{y}{4 x} \geqslant 3 \sqrt[3]{x^{2} \cdot \frac{y}{4 x} \cdot \frac{y}{4 x}}=3 \sqrt[3]{\frac{y^{2}}{16}}$, so $y^{2} \leqslant \frac{16}{27}$, which means when $\sin \frac{\theta}{2}=\frac{\sqrt{3}}{3}$, $y_{\text {max }}=\frac{4 \sqrt{3}}{9}$.
It should be noted that when using the AM-GM inequality to find the maximum value, the following must be observed:
（1）All terms or factors must be non-negative;
（2）The sum or product is a constant;
（3）All terms or factors can be equal.
If necessary, appropriate transformations should be made to meet the above conditions.