If $\cos \left( {\alpha – \beta } \right) + \cos \left( {\beta – \gamma } \right) + \cos \left( {\gamma – \alpha } \right) = – \frac{3}{2}$, where $(α,β,γ ∈ R).$
(A) $\sum {\cos \alpha } = 0$
(B) $\sum {\sin \alpha } = 0$
(C) $\sin \alpha \sin \beta \sin \gamma = 0$
(D) $\sum {\cos \alpha } +\sum {\sin \alpha } = 0$
This is a multiple choice question with one or more options
My approach is bases on Complex number
$\cos \left( {\alpha – \beta } \right) + \cos \left( {\beta – \gamma } \right) + \cos \left( {\gamma – \alpha } \right) = – \frac{3}{2}$
$T = {e^{i\left( {\alpha – \beta } \right)}} + {e^{i\left( {\beta – \gamma } \right)}} + {e^{i\left( {\gamma – \alpha } \right)}}$
${\mathop{\rm Re}\nolimits} \left( T \right) = – \frac{3}{2}$
${e^{i\frac{\alpha }{\beta }}} + {e^{i\frac{\beta }{\gamma }}} + {e^{i\frac{\gamma }{\alpha }}} \Rightarrow {e^{i\frac{\alpha }{\beta }}}\left( {1 + \frac{{{e^{i\frac{\beta }{\gamma }}}}}{{{e^{i\frac{\alpha }{\beta }}}}}} \right) + {e^{i\frac{\gamma }{\alpha }}} \Rightarrow {e^{i\frac{\alpha }{\beta }}}\left( {1 + {e^{i\left( {\frac{\beta }{\gamma } = \frac{\alpha }{\beta }} \right)}}} \right) + {e^{i\frac{\gamma }{\alpha }}}$
How do I proceed from here
Best Answer
Note that $\cos (\alpha -\beta)=\cos \alpha \cos \beta +\sin \alpha \sin \beta$, therefore:
$$(\sum \sin \alpha)^2+(\sum \cos \alpha)^2=3+2(\sum \sin \alpha \sin \beta+\sum \cos \alpha \cos \beta)=3+2(\frac{-3}{2})=0;$$
$$\implies \sum \sin \alpha=\sum \cos \alpha=0.$$