Question

If $0<\alpha ,\beta <\pi$ and $\cos \alpha +\cos \beta -\cos (\alpha +\beta )=\frac{3}{2}$, then

• A. $\alpha =\frac{\pi }{3}$
• B. $\beta =\frac{\pi }{3}$
• C. $\alpha =\beta$
• D. $\alpha +\beta =\frac{\pi }{3}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\alpha =\frac{\pi }{3}$
B $\alpha =\beta$
C $\beta =\frac{\pi }{3}$

$\cos +\cos \beta -\cos (\alpha +\beta )=\frac{3}{2}\Rightarrow 2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}-(2{\cos }^2\frac{\alpha +\beta }{2}-1)=\frac{3}{2}\Rightarrow 2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}-2{\cos }^2\frac{\alpha +\beta }{2}=\frac{3}{2}-1\Rightarrow 4{\cos }^2\frac{\alpha +\beta }{2}-4\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}+1=0\Rightarrow 4{\cos }^2\frac{\alpha +\beta }{2}-4\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}+{\cos }^2\frac{\alpha -\beta }{2}+{\sin }^2\frac{\alpha -\beta }{2}=0\Rightarrow {(2\cos \frac{\alpha +\beta }{2}-\cos \frac{\alpha -\beta }{2})}^2+{\sin }^2\frac{\alpha -\beta }{2}=0$
$\Rightarrow {(2\cos \frac{\alpha +\beta }{2}-\cos \frac{\alpha -\beta }{2})}^2=0$ or ${\sin }^2\frac{\alpha -\beta }{2}=0$
$\Rightarrow 2\cos \frac{\alpha +\beta }{2}-\cos \frac{\alpha -\beta }{2}=0$ or $\sin \frac{\alpha -\beta }{2}=0$
From $\sin \frac{\alpha -\beta }{2}=0\Rightarrow \frac{\alpha -\beta }{2}=0\Rightarrow \alpha -\beta =0\Rightarrow \alpha =\beta$ ..(1)
And from
$2\cos \frac{\alpha +\beta }{2}-\cos \frac{\alpha -\beta }{2}=0\Rightarrow 2\cos \left(\frac{\alpha }{2}\right)\cos \left(\frac{\beta }{2}\right)-2\sin \left(\frac{\alpha }{2}\right)\sin \left(\frac{\beta }{2}\right)-\cos \left(\frac{\alpha }{2}\right)\cos \left(\frac{\beta }{2}\right)-\sin \left(\frac{\alpha }{2}\right)\sin \left(\frac{\beta }{2}\right)=0\Rightarrow \tan \left(\frac{\alpha }{2}\right)\tan \left(\frac{\beta }{2}\right)=\frac{1}{3}$
U\sin g (1)
${\tan }^2\left(\frac{\alpha }{2}\right)=3\Rightarrow \tan \frac{\alpha }{2}=\frac{1}{\sqrt{3}}\Rightarrow \alpha =\frac{\pi }{3}=\beta$