Question Number 20646 by oyshi last updated on 30/Aug/17
$${if}\:\mathrm{cos}\:\left(\beta−\gamma\right)+\mathrm{cos}\left(\:\gamma−\alpha\right)+\mathrm{cos}\:\left(\alpha−\beta\right)=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${so}\:{proof}\:{it}, \\ $$$$\Sigma\mathrm{cos}\:\alpha=\mathrm{0},\Sigma\mathrm{sin}\:\alpha=\mathrm{0} \\ $$
Answered by $@ty@m last updated on 31/Aug/17
$${Given}, \\ $$$$\mathrm{cos}\:\left(\beta−\gamma\right)+\mathrm{cos}\left(\:\gamma−\alpha\right)+\mathrm{cos}\:\left(\alpha−\beta\right)=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2cos}\:\left(\beta−\gamma\right)+\mathrm{2cos}\left(\:\gamma−\alpha\right)+\mathrm{2cos}\:\left(\alpha−\beta\right)=−\mathrm{3} \\ $$$$\Rightarrow\Sigma\left(\mathrm{2}{cos}\beta{cos}\gamma+\mathrm{2}{sin}\beta{sin}\gamma\right)\:+\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\Sigma\left(\mathrm{2}{cos}\beta{cos}\gamma+\mathrm{2}{sin}\beta{sin}\gamma\right)\:+\Sigma\left({sin}^{\mathrm{2}} \alpha+{cos}^{\mathrm{2}} \alpha\right)=\mathrm{0} \\ $$$$\Rightarrow\left\{\Sigma{cos}^{\mathrm{2}} \alpha+\Sigma\mathrm{2}{cos}\beta{cos}\gamma\right\}+\left\{\Sigma{sin}^{\mathrm{2}} \alpha+\Sigma\mathrm{2}{sin}\beta{sin}\gamma\right\}=\mathrm{0} \\ $$$$\Rightarrow\left(\Sigma\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +\left(\Sigma\mathrm{sin}\:\alpha\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\Sigma\mathrm{cos}\:\alpha=\mathrm{0},\Sigma\mathrm{sin}\:\alpha=\mathrm{0} \\ $$
Commented by ajfour last updated on 31/Aug/17
$${thanks}\:{immensely}. \\ $$