Question

Let $\alpha$ and $\beta$ be such that $\pi <\alpha -\beta <2\pi$. If $\sin \alpha +\sin \beta =-\frac{21}{65}$ and $\cos \alpha +\cos \beta =\frac{17}{65}$, then the value of $\cos \frac{\alpha -\beta }{2}$ is

• A. $\frac{4}{\sqrt{130}}$
• B. $\frac{3}{\sqrt{130}}$
• C. $\frac{6}{65}$
• D. None of these
  

Answer 1

Answer: (D)

None of these

$\pi <\alpha -\beta <2\pi \Rightarrow \frac{\pi }{2}<\frac{\alpha -\beta }{2}<\pi \Rightarrow \cos \left(\frac{\alpha -\beta }{2}\right)<0$
$\sin \alpha +\sin \beta =-\frac{21}{65}$ ...(1)
$\cos \alpha +\cos \beta =\frac{17}{65}$ ...(2)
Squaring and adding (1) & (2)
$1+1+2(\sin \alpha \sin \beta +\cos \alpha \cos \beta )={\left(\frac{-21}{65}\right)}^2+{\left(\frac{17}{65}\right)}^2\Rightarrow 2(1+\cos (\alpha -\beta ))={\left(\frac{-21}{65}\right)}^2+{\left(\frac{17}{65}\right)}^2\Rightarrow {\cos }^2\left(\frac{\alpha -\beta }{2}\right)=\frac{730}{4{\left(65\right)}^2}$
$\Rightarrow \cos \left(\frac{\alpha -\beta }{2}\right)=-\frac{\sqrt{730}}{130}$ ...{ $\cos \left(\frac{\alpha -\beta }{2}\right)<0$}