Question

If $\sin \alpha +\sin \beta =\frac{1}{2}$ and $\cos \alpha +\cos \beta =\frac{\sqrt{3}}{2}$, then value of $3\alpha +\beta$ is

• A. ${90}^0$
• B. $0^0$
• C. ${120}^0$
• D. ${60}^0$

Answer 1

The correct option is B $0^0$

from first equation we get
$2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}=\frac{1}{2}$
from second equation we get
$2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}=\frac{\sqrt{3}}{2}$
we divide the 2 equations
$\tan \frac{\alpha +\beta }{2}=\frac{1}{\sqrt{3}}$
so$\frac{\alpha +\beta }{2}=\frac{\pi }{6}$
so $\alpha +\beta =\frac{\pi }{3}$ .....(1)
now we square the 2 original equations and add them
$1+1+2\cos (\beta -\alpha )=1$
$\cos (\beta -\alpha )=-\frac{1}{2}$
so $\beta -\alpha =\frac{2\pi }{3}$.....(2)
from (1) and (2) we get
$\alpha =-\frac{\pi }{6}$ and $\beta =\frac{\pi }{2}$
so $3\alpha +\beta =0$