Let $\alpha ,\beta$  be such that $\pi <\alpha -\beta <3\pi$  If $\sin \alpha +\sin \beta =-\frac{21}{65}$  and $\cos \alpha +\cos \beta =-\frac{27}{65}$   then $\cos \left(\frac{\alpha -\beta }{2}\right)=$

$\pi <\alpha -\beta <3\pi \Rightarrow \frac{\pi }{2}<\frac{\alpha -\beta }{2}<\frac{3\pi }{2}$ 

${(\sin \alpha +\sin \beta )}^2+{(\cos \alpha +\cos \beta )}^2=\frac{{21}^2}{{65}^2}+\frac{{27}^2}{{65}^2}$

$2+2\cos \left(\alpha -\beta \right)=\frac{9}{{65}^2}\left(49+81\right)=\frac{9}{65\times 65}\times 130=\frac{18}{65}$

$1+\cos \left(\alpha -\beta \right)=\frac{9}{65}\Rightarrow 2{\cos }^2\left(\frac{\alpha -\beta }{2}\right)=\frac{9}{65}\Rightarrow {\cos }^2\left(\frac{\alpha -\beta }{2}\right)=\frac{9}{130}$

$\cos \left(\frac{\alpha -\beta }{2}\right)=-\sqrt{\frac{9}{130}}=-\frac{3}{\sqrt{130}}$