Question Number 211989 by Spillover last updated on 26/Sep/24
Answered by som(math1967) last updated on 26/Sep/24
$$\:{cos}\alpha{cos}\beta+{sin}\alpha{sin}\beta+{cos}\beta{cos}\gamma \\ $$$$+{sin}\beta{sin}\gamma+{cos}\gamma{cos}\alpha+{sin}\gamma{sin}\alpha \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{2}\left({cos}\alpha{cos}\beta+{cos}\beta{cos}\gamma+{cos}\gamma{cos}\alpha\right) \\ $$$$+\mathrm{2}\left({sin}\alpha{sin}\beta+{sin}\beta{sin}\gamma+{sin}\alpha{sin}\gamma\right) \\ $$$$=−\mathrm{1}−\mathrm{1}−\mathrm{1} \\ $$$$\:{cos}^{\mathrm{2}} \alpha+{cos}^{\mathrm{2}} \beta+{cos}^{\mathrm{2}} \gamma \\ $$$$+\mathrm{2}\left({cos}\alpha{cos}\beta+{cos}\beta{cos}\gamma+{cos}\gamma{cos}\alpha\right) \\ $$$$+{sin}^{\mathrm{2}} \alpha+{sin}^{\mathrm{2}} \beta+{sin}^{\mathrm{2}} \gamma \\ $$$$+\mathrm{2}\left({sin}\alpha{sin}\beta+{sin}\beta{sin}\gamma+{sin}\alpha{sin}\gamma\right)=\mathrm{0} \\ $$$$\left({cos}\alpha+{cos}\beta+{cos}\gamma\right)^{\mathrm{2}} +\left({sin}\alpha+{sin}\beta+{sin}\gamma\right)^{\mathrm{2}} \\ $$$$=\mathrm{0} \\ $$$$\left.\therefore{cos}\alpha+{cos}\beta+{cos}\gamma={sin}\alpha+{sin}\beta+{sin}\gamma\right)=\mathrm{0} \\ $$