Question

Let $\alpha ,\beta$ be such that $0<\alpha -\beta -\pi <2\pi .$ 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 a

• A. $\frac{6}{65}$
• B. $\frac{3}{\sqrt{130}}$
• C. $\frac{-3}{\sqrt{130}}$
• D. $\frac{-6}{65}$

Answer 1

Answer: (C)

$\frac{-3}{\sqrt{130}}$

${(\sin \alpha +\sin \beta )}^2+{(\cos \alpha +\cos \beta )}^2$
$\Rightarrow 2+2\cos (\alpha -\beta )=\frac{18}{65}$
$\Rightarrow {\cos }^2\frac{\alpha -\beta }{2}=\frac{9}{130},As\pi <\alpha -\beta <3\pi \Rightarrow \frac{\pi }{2}<\frac{\alpha -\beta }{2}<\frac{3\pi }{2}$
$\Rightarrow \cos \frac{\alpha -\beta }{2}=\frac{-3}{\sqrt{130}}$