15. Let $0<\theta<\pi$, find the maximum value of $\sin \frac{\theta}{2}(1+\cos \theta)$.
15. Let $y=\sin \frac{\theta}{2}(1+\cos \theta)=2 \sin \frac{\theta}{2} \cos ^{2} \frac{\theta}{2}$, then $y^{2}=4 \sin ^{2} \frac{\theta}{2} \cos ^{2} \frac{\theta}{2} \cos ^{2} \frac{\theta}{2}=$ $2\left[2 \sin ^{2} \frac{\theta}{2}\right] \cos ^{2} \frac{\theta}{2} \cdot \cos ^{2} \frac{\theta}{2} \leqslant 2\left[\frac{2}{3}\right]^{3}=\frac{16}{27}$, so $-\frac{4 \sqrt{3}}{9} \leqslant y \leqslant \frac{4 \sqrt{3}}{9}$. When $\theta \in(0, \pi)$, hence $y>0$, and $y$ achieves its maximum value of $\frac{4 \sqrt{3}}{9}$ if and only if $2 \sin ^{2} \frac{\theta}{2}=\cos ^{2} \frac{\theta}{2}, \tan ^{2} \frac{\theta}{2}=\frac{1}{2}, \cos \theta=\frac{1}{3}$.