Mathematics - Trigonometric Equation Question with Solution | TestHub

Given
$\cos \alpha + \cos \beta = \frac{3}{2}$
$\sin \alpha + \sin \beta = \frac{1}{2}$
$\Rightarrow 2\mathrm{c}\mathrm{o}\mathrm{s}\left(\frac{\alpha +\beta }{2}\right)\mathrm{c}\mathrm{o}\mathrm{s}\left(\frac{\alpha -\beta }{2}\right)=\frac{3}{2}$......(i) and

$2\sin \left(\frac{\alpha +\beta }{2}\right)\mathrm{c}\mathrm{o}\mathrm{s}(\frac{\alpha -\beta }{2})=\frac{1}{2}$....(ii)

Dividing (ii) and (i)

$\tan \left(\frac{\left(\alpha + \beta \right)}{2}\right)$
$\therefore \tan \mathrm{\theta } =\frac{1}{3}$

$\sin 2\mathrm{\theta }+\cos 2\mathrm{\theta } =$$\frac{2\tan \mathrm{\theta }}{1+{\tan }^2\mathrm{\theta }}+ \frac{1-{\tan }^2\mathrm{\theta }}{1+{\tan }^2\mathrm{\theta }}=\frac{2\left(\frac{1}{3}\right)}{1+\frac{1}{9}}+\frac{1-\frac{1}{9}}{1+\frac{1}{9}}$

$= \frac{\frac{2}{3}+\frac{8}{9}}{\frac{10}{9}}=\frac{14}{10}=\frac{7}{5}$