Let $\alpha, \beta$ be such that $\pi < \alpha-\beta < 3 \pi$. If $\sin \alpha+\sin \beta=-\frac{21}{65}$…
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 the value of $\cos \frac{\alpha-\beta}{2}$ is
$-\frac{3}{\sqrt{130}}$
$\frac{3}{\sqrt{130}}$
$\frac{6}{65}$
$-\frac{6}{65}$
Solution
$\sin \alpha+\sin \beta=-\frac{21}{65}$ and $\cos \alpha+\cos \beta=-\frac{27}{65}$.
Squaring and adding, we get
$
\begin{aligned}
& 2+2 \cos (\alpha-\beta)=\frac{1170}{(65)^2} \\
& \Rightarrow \cos ^2\left(\frac{\alpha-\beta}{2}\right)=\frac{9}{130} \Rightarrow \cos \left(\frac{\alpha-\beta}{2}\right)=\frac{-3}{\sqrt{130}} \quad\left(\because \frac{\pi}{2} < \frac{\alpha-\beta}{2} < \frac{3 \pi}{2}\right) .
\end{aligned}
$