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

Answer: (A)

Given that, $\sin \alpha +\sin \beta =-\frac{21}{65}$ ...(i) and $\cos \alpha +\cos \beta =-\frac{27}{65}$ ...(ii) Squaring Eqs. (i) and (ii) and then adding, we get ${(\sin \alpha +\sin \beta )}^2+{(\cos \alpha +\cos \beta )}^2$ $={\left(-\frac{21}{65}\right)}^2+{\left(-\frac{27}{65}\right)}^2$ ${\sin }^2\alpha +{\sin }^2\beta +2\sin \alpha \sin \beta +{\cos }^2\alpha +{\cos }^2\beta$ $+2\cos \alpha \cos \beta =\frac{1170}{4225}$ $\Rightarrow$ $2+2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )=\frac{1170}{4225}$ $\Rightarrow$ $2+2\cos (\alpha -\beta )=\frac{1170}{4225}$ $\Rightarrow$ $2[1+\cos (\alpha -\beta )]=\frac{1170}{4225}$ $\Rightarrow$ $2\left[2{\cos }^2\left(\frac{\alpha -\beta }{2}\right)\right]=\frac{1170}{4225}$ $\Rightarrow$ ${\cos }^2\left(\frac{\alpha -\beta }{2}\right)=\frac{1170}{4\times 4225}$ $\Rightarrow$ ${\cos }^2\left(\frac{\alpha -\beta }{2}\right)=\frac{9}{130}$ $\Rightarrow$ $\cos \left(\frac{\alpha -\beta }{2}\right)=-\frac{3}{\sqrt{130}}$ $(\because \pi <\alpha -\beta <3\pi )$