Given $sin{\mkern 1mu} \alpha + \sin \beta = - {{21} \over {65}}$ .........(1) and $\cos \alpha + \cos \beta = - {{27} \over {65}}$ ........(2) Square and add (1) and (2) you will get 2\left( {1 + \cos \alpha \cos \beta + \sin \alpha \sin \beta } \right)$$$$ = {{{{\left( {21} \right)}^2} + {{\left( {27} \right)}^2}} \over {{{\left( {65} \right)}^2}}} $\Rightarrow$ $2\left( {1 + \cos \left( {\alpha - \beta } \right)} \right) = {{1170} \over {{{\left( {65} \right)}^2}}}$ $\Rightarrow$ 4{\cos ^2}{{\alpha - \beta } \over 2}$$$$ = {{1170} \over {{{\left( {65} \right)}^2}}} $\Rightarrow$ {\cos ^2}{{\alpha - \beta } \over 2}$$$$ = {9 \over {130}} $\therefore$ $\cos {{\alpha - \beta } \over 2} = \pm {3 \over {\sqrt {130} }}$ [ But $\cos {{\alpha - \beta } \over 2} \ne + {3 \over {\sqrt {130} }}$ as $\pi < \alpha - \beta < 3\pi$ $\Rightarrow$ ${\pi \over 2} < {{\alpha - \beta } \over 2} < {{3\pi } \over 2}$ $\Rightarrow$ $\cos {{\alpha - \beta } \over 2} < 0$ ] So $\cos {{\alpha - \beta } \over 2} = - {3 \over {\sqrt {130} }}$