3.365. $\cos \frac{\alpha-\beta}{2}$, if $\sin \alpha+\sin \beta=-\frac{27}{65} ; \operatorname{tg} \frac{\alpha+\beta}{2}=\frac{7}{9} ; \frac{5}{2} \pi<\alpha<3 \pi$ and $-\frac{\pi}{2}<\beta<0$.
## Solution.

$$
\begin{aligned}
& \sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} = -\frac{27}{65} \\
& \operatorname{tg} \frac{\alpha + \beta}{2} = \frac{7}{9} \Rightarrow \frac{\sin \frac{\alpha + \beta}{2}}{\cos \frac{\alpha + \beta}{2}} = \frac{7}{9} \Rightarrow \frac{\sin^2 \frac{\alpha + \beta}{2}}{\cos^2 \frac{\alpha + \beta}{2}} = \frac{49}{81} \\
& \frac{\sin^2 \frac{\alpha + \beta}{2}}{1 - \sin^2 \frac{\alpha + \beta}{2}} = \frac{49}{81} \Rightarrow \sin \frac{\alpha + \beta}{2} = \pm \frac{7}{\sqrt{130}}
\end{aligned}
$$

Since $\frac{5}{2} \pi < \alpha < 3 \pi$ and $-\frac{\pi}{2} < \beta < 0$, then $\pi < \frac{\alpha + \beta}{2} < \frac{3 \pi}{2}$, i.e., the angle $\frac{\alpha + \beta}{2}$ is in the third quadrant of the unit circle and $\sin \frac{\alpha + \beta}{2} < 0$.

$$
\begin{aligned}
& \text{Therefore, } \sin \frac{\alpha + \beta}{2} = -\frac{7}{\sqrt{130}} \Rightarrow -\frac{14}{\sqrt{130}} \cos \frac{\alpha - \beta}{2} = -\frac{27}{65} \Rightarrow \\
& \Rightarrow \cos \frac{\alpha - \beta}{2} = \frac{27 \sqrt{130}}{14 \cdot 65} = \frac{27 \sqrt{130} \cdot \sqrt{130}}{14 \cdot 65 \cdot \sqrt{130}} = \frac{27 \cdot 130}{14 \cdot 65 \cdot \sqrt{130}} = \frac{27}{7 \sqrt{130}}
\end{aligned}
$$

$$
\text{Answer: } \cos \frac{\alpha - \beta}{2} = \frac{27}{7 \sqrt{130}}
$$