Question

Let $A$ and $B$ denote the statements $A:\cos \alpha +\cos \beta +\cos \gamma =0$, $B:\sin \alpha +\sin \beta +\sin \gamma =0$

If $\cos (\beta -\gamma )+\cos (\gamma -\alpha )+\cos (\alpha -\beta )=-\frac{3}{2}$, then

• A. $A$ is false and $B$ is true
• B. both $A$ and $B$ are true
• C. both $A$ and $B$ are false
• D. $A$ istrue and $B$ is false

Answer 1

Answer: (B)

both $A$ and $B$ are true

$\cos (\beta -\gamma )+\cos (\alpha -\beta )+\cos (\gamma -\alpha )=-\frac{3}{2}$

$\Rightarrow \cos \beta \cos \gamma +\sin \beta \sin \gamma +\cos \alpha \cos \beta +\sin \alpha \sin \beta +\cos \gamma \cos \alpha +\sin \gamma \sin \alpha =-\frac{3}{2}$

$\Rightarrow (2\cos \beta \cos \gamma +2\cos \alpha \cos \beta +2\cos \gamma \cos \alpha )+(2\sin \beta \sin \gamma +2\sin \alpha \sin \beta +2\sin \gamma \sin \alpha )=-3$

$\Rightarrow 3+(2\cos \beta \cos \gamma +2\cos \alpha \cos \beta +2\cos \gamma \cos \alpha )+(2\sin \beta \sin \gamma +2\sin \alpha \sin \beta +2\sin \gamma \sin \alpha )=-3+3$

$\Rightarrow {\sin }^2\alpha +{\cos }^2\alpha +{\sin }^2\beta +{\cos }^2\beta +{\sin }^2\gamma +{\cos }^2\gamma +(2\cos \beta \cos \gamma +2\cos \alpha \cos \beta +2\cos \gamma \cos \alpha )+(2\sin \beta \sin \gamma +2\sin \alpha \sin \beta +2\sin \gamma \sin \alpha )=0$

$\Rightarrow ({\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma +2\cos \beta \cos \gamma +2\cos \alpha \cos \beta +2\cos \gamma \cos \alpha )+({\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma +2\sin \beta \sin \gamma +2\sin \alpha \sin \beta +2\sin \gamma \sin \alpha )=0{\Rightarrow (\cos \alpha +\cos \beta +\cos \gamma )}^2+{(\sin \alpha +\sin \beta +\sin \gamma )}^2=0$

This is only true when,

$\cos \alpha +\cos \beta +\cos \gamma =0$

and $\sin \alpha +\sin \beta +\sin \gamma =0$