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 the value of $\cos \frac{\alpha -\beta }{2}$is:

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

Answer 1

The correct option is A $-\frac{3}{\sqrt{130}}$

Given $\sin \alpha +\sin \beta =-\frac{21}{65}$ (i) and $\cos \alpha +\cos \beta =-\frac{27}{65}$ (ii)
Squaring and adding both the equation we get,
$2+2\cos (\alpha -\beta )=\frac{{21}^2+{27}^2}{{65}^2}=\frac{18}{65}$
$\Rightarrow 2.2{\cos }^2\left(\frac{\alpha -\beta }{2}\right)=\frac{18}{65}$
$\Rightarrow \cos \left(\frac{\alpha -\beta }{2}\right)=\pm \sqrt{\frac{9}{130}}$
But given $\pi <\alpha -\beta <3\pi$ $\therefore \cos \left(\frac{\alpha -\beta }{2}\right)=-\sqrt{\frac{9}{130}}=-\frac{3}{\sqrt{130}}$