Question

If $\sin \alpha +\sin \beta +\sin \gamma =0$ and $\cos \alpha +\cos \beta +\cos \gamma =0$, then value of $\cos (\alpha -\beta )+\cos (\beta -\gamma )+\cos (\gamma +\alpha )$ is

• A. $-\frac{3}{2}$
• B. $-1$
• C. $-\frac{1}{2}$
• D. $0$

Answer 1

Answer: (A)

$-\frac{3}{2}$

$\sin \alpha +\sin \beta +\sin \gamma =0\Rightarrow \sin \alpha +\sin \beta =-\sin \gamma$ ...(1)
$\cos \alpha +\cos \beta +\cos \gamma =0\Rightarrow \cos \alpha +\cos \beta =-\cos \gamma$ ...(2)
Squaring and adding (1) and (2), we get
${(\sin \alpha +\sin \beta )}^2+{(\cos \alpha +\cos \beta )}^2={\sin }^2\gamma +{\cos }^2\gamma \Rightarrow 2+2\cos (\alpha -\beta )=1\Rightarrow \cos (\alpha -\beta )=-\frac{1}{2}$
Similarly, $\cos (\beta -\gamma )=-\frac{1}{2},\cos (\gamma -\alpha )=-\frac{1}{2}$
Therefore,
$\cos (\alpha -\beta )+\cos (\beta -\gamma )+\cos (\gamma -\alpha )=-\frac{3}{2}$