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

Squaring and adding the given relations
$1+1+2\cos (\alpha -\beta )=\frac{441+729}{65\times 65}$
or $2[1+\cos (\alpha -\beta )]=\frac{1170}{65\times 65}=\frac{13\times 5\times 18}{65\times 65}$
or $2.2{\cos }^2\frac{\alpha -\beta }{2}=\frac{18}{65}=\frac{36}{130}$
$\therefore \cos \frac{\alpha -\beta }{2}=\pm \frac{3}{\sqrt{130}}$
Since $\pi <\alpha -\beta <3\pi \Rightarrow \frac{\pi }{2}<\frac{\alpha -\beta }{2}<\frac{3\pi }{2}$
in the above interval $\cos \frac{\alpha -\beta }{2}$ is -ive
$\therefore \cos \frac{\alpha -\beta }{2}=-\frac{3}{\sqrt{130}}$