16. Given $\sin \alpha+\cos \alpha=-\frac{3}{\sqrt{5}}$, and $|\sin \alpha|>|\cos \alpha|$, find the value of $\tan \frac{\alpha}{2}$.
16. From the problem, we get $\sin \alpha < 0$, and $|\sin \alpha| > |\cos \alpha|$, so $\sin \alpha < \cos \alpha$. From $\sin \alpha + \cos \alpha = -\frac{3}{\sqrt{5}}$, we get $\sin \alpha - \cos \alpha = -\frac{1}{\sqrt{5}}$, thus $\sin \alpha = -\frac{2}{\sqrt{5}}, \cos \alpha = -\frac{1}{\sqrt{5}}$, hence $\tan \frac{\alpha}{2} = \frac{1 - \cos \alpha}{\sin \alpha} = -\frac{\sqrt{5} + 1}{2}$.