Question

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

• A. $\frac{3}{5}$
• B. $\frac{4}{5}$
• C. $\frac{7}{5}$
• D. $\frac{8}{5}$

Answer 1

The correct option is C $\frac{7}{5}$

$\cos \alpha +\cos \beta =\frac{3}{2}$

$\Rightarrow$ $2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)=\frac{3}{2}$ ---- ( 1 )

$\sin \alpha +\sin \beta =\frac{1}{2}$

$\Rightarrow$ $2\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)=\frac{1}{2}$ ---- ( 2 )

Dividing ( 2 ) by ( 1 ) we get,

$\Rightarrow$ $\tan \left(\frac{\alpha +\beta }{2}\right)=\frac{1}{3}$ ----- ( 3 )

It is given that $\theta =\frac{\alpha +\beta }{2}$

$\Rightarrow$ $\sin 2\theta +\cos 2\theta$

$\Rightarrow$ $\frac{2\tan \left(\frac{\alpha +\beta }{2}\right)}{1+{\tan }^2\left(\frac{\alpha +\beta }{2}\right)}$ $+\frac{1-{\tan }^2\left(\frac{\alpha +\beta }{2}\right)}{1+{\tan }^2\left(\frac{\alpha +\beta }{2}\right)}$

$\Rightarrow$ $\frac{\frac{2}{3}}{1+\frac{1}{9}}+\frac{1-\frac{1}{9}}{1+\frac{1}{9}}$ [ From ( 3 ) ]

$\Rightarrow$ $\frac{\frac{2}{3}}{\frac{10}{9}}+\frac{\frac{8}{9}}{\frac{10}{9}}$

$\Rightarrow$ $\frac{6}{10}+\frac{8}{10}$

$\Rightarrow$ $\frac{14}{10}$

$\Rightarrow$ $\frac{7}{5}$