question_answer

If \[\cos \alpha +\cos \beta =\frac{3}{2}\]and \[\sin \alpha +\sin \beta =\frac{1}{2}\]and \[\theta \]is the arithmetic mean of \[\alpha \] and \[\beta ,\] then \[\sin 2\theta +\cos 2\theta \]is equal to: [JEE Main Online Paper (Held On 11 April 2015)]  

A) \[\frac{3}{5}\]

B) \[\frac{4}{5}\]

C) \[\frac{7}{5}\]

D) \[\frac{8}{5}\]

Correct Answer: C

Solution :

  \[2\cos \left( \frac{\alpha +\beta }{2} \right)\cos \left( \frac{\alpha -\beta }{2} \right)=\frac{3}{2}\] ?(1) \[2\sin \left( \frac{\alpha +\beta }{2} \right)\cos \left( \frac{\alpha -\beta }{2} \right)=\frac{1}{2}\] ?.(2) Dividing (2) by (1) , \[\tan \left( \frac{\alpha +\beta }{2} \right)=\frac{1}{3}\] \[\Rightarrow \]\[\tan \theta =\frac{1}{3}\left( \because \theta =\frac{\alpha +\beta }{2}\text{given} \right)\] \[\Rightarrow \]\[\sin \theta =\frac{1}{\sqrt{10}}\]\[\And \cos \theta =\frac{3}{\sqrt{10}}\] \[\sin 2\theta +\cos 2\theta =2\sin \theta +2{{\cos }^{2}}\theta -1\] \[=2\times \frac{1}{\sqrt{10}}\times \frac{3}{\sqrt{10}}+2\left( \frac{9}{10} \right)-1\] \[=\frac{6}{10}+\frac{18}{10}-1\] \[=\frac{7}{5}\]