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

Answer 1

Answer: (B)

$\frac{7}{5}$

Given $\theta$ is the arithmetic mean of $\alpha$ and $\beta$
$\Rightarrow \theta =\frac{\alpha +\beta }{2}$
Also given $\cos \alpha +\cos \beta =\frac{3}{2}$
$\Rightarrow 2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)=\frac{3}{2}$
$\Rightarrow 2\cos \theta \cos \left(\frac{\alpha -\beta }{2}\right)=\frac{3}{2}$ .....(1)
$\sin \alpha +\sin \beta =\frac{1}{2}$
$\Rightarrow 2\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)=\frac{1}{2}$
$\Rightarrow 2\sin \theta \cos \left(\frac{\alpha -\beta }{2}\right)=\frac{1}{2}$ .....(2)
Dividing (1) by (2), we get
$\cot \theta =3$
$\cos \theta =\frac{3}{\sqrt{10}}$
$\sin \theta =\frac{1}{\sqrt{10}}$
Now, $\sin 2\theta +\cos 2\theta =2\sin \theta \cos \theta +1-2{\sin }^2\theta$
$=\frac{6}{10}+1-\frac{2}{10}$
$=\frac{7}{5}$ (Since maximum value $\sin 2\theta +\cos 2\theta$ is $2$)