Question

Let $\alpha$ and $\beta$ 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, $\sin \alpha +\sin \beta =-\frac{21}{65}$

and $\cos \alpha +\cos \beta =-\frac{27}{65}$

On squaring and adding the above equations, we get

${\sin }^2\alpha \cdot {\sin }^2\beta +2\sin \alpha \sin \beta +{\cos }^2\alpha +{\cos }^2\beta +2\cos \alpha \cos \beta$

$={\left(\frac{21}{65}\right)}^2+{\left(\frac{27}{65}\right)}^2$

$\Rightarrow 2+2(\sin \alpha \cdot \sin \beta +\cos \alpha \cos \beta )=\frac{1170}{4225}$

$\Rightarrow 2+2\cos (\alpha -\beta )=\frac{1170}{4225}$

$\Rightarrow 2[1+\cos (\alpha -\beta )]=\frac{1170}{4225}$

$\Rightarrow 2\left[2{\cos }^2\left(\frac{\alpha -\beta }{2}\right)\right]=\frac{1170}{4225}$

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

$\Rightarrow \cos \left(\frac{\alpha -\beta }{2}\right)=-\frac{3}{\sqrt{130}}$ ....$(\because \pi <\alpha -\beta <3\pi )$