Question

$\pi <\alpha -\beta <3\pi ,\sin \alpha +\sin \beta =\frac{-21}{65},$ $\cos \alpha +\cos \beta =\frac{-27}{65}$. The value of $\cos \left(\frac{\alpha -\beta }{2}\right)$ is

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

Answer 1

Answer: (B)

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

$\sin \alpha +\sin \beta =-\frac{21}{65},\cos \alpha +\cos \beta =-\frac{27}{65}$
On squaring and adding, we get
$2+2\cos (\alpha -\beta )=\frac{441}{4225}+\frac{729}{4225}=\frac{1170}{4225}$

$1+\cos (\alpha -\beta )=\frac{585}{4225}$

$1+2{\cos }^2\left(\frac{\alpha -\beta }{2}\right)-1=\frac{585}{4225}$

${\cos }^2\left(\frac{\alpha -\beta }{2}\right)=\frac{9}{130}$

As $\pi <(\alpha -\beta )<3\pi \Rightarrow \frac{\pi }{2}<\frac{(\alpha -\beta )}{2}<\frac{3\pi }{2}$

$\therefore \cos \left(\frac{\alpha -\beta }{2}\right)=\frac{-3}{\sqrt{130}}$