Question

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 value of $\cos \frac{\alpha -\beta }{2}$ equals

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

Answer 1

Answer: (C)

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

Given $\sin \alpha +\sin \beta =-\frac{21}{65}$ and $\cos \alpha +\cos \beta =-\frac{27}{65}$
Squaring and adding ,we get
$2(1+\cos \alpha \cos \beta +\sin \alpha \sin \beta )=\frac{1170}{{\left(65\right)}^2}$
$2[1+\cos (\alpha -\beta )]=\frac{1170}{{\left(65\right)}^2}$
${\cos }^2\frac{\alpha -\beta }{2}=\frac{1170}{4\times 65\times 65}=\frac{130\times 9}{\left(130\right)\times \left(130\right)}$
${\cos }^2\frac{\alpha -\beta }{2}=\frac{9}{130}$
$\therefore \cos \frac{\alpha -\beta }{2}=\pm \frac{3}{\sqrt{130}}$
Since, $\pi <\alpha -\beta <3\pi$
$\frac{\pi }{2}<\frac{\alpha -\beta }{2}<\frac{3\pi }{2}$
So, $\cos \left(\frac{\alpha -\beta }{2}\right)=\frac{-3}{\sqrt{130}}$