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 ) $