Question Number 18062 by Tinkutara last updated on 14/Jul/17
$$\mathrm{If}\:\mathrm{0}\:<\:\alpha,\:\beta\:<\:\pi\:\mathrm{and}\:\mathrm{they}\:\mathrm{satisfy} \\ $$$$\mathrm{cos}\:\alpha\:+\:\mathrm{cos}\:\beta\:−\:\mathrm{cos}\:\left(\alpha\:+\:\beta\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\:\alpha\:=\:\beta \\ $$$$\left(\mathrm{2}\right)\:\alpha\:+\:\beta\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\left(\mathrm{3}\right)\:\alpha\:=\:\mathrm{2}\beta \\ $$$$\left(\mathrm{4}\right)\:\beta\:=\:\mathrm{2}\alpha \\ $$
Answered by Tinkutara last updated on 16/Jul/17
$$\mathrm{2}\:\mathrm{cos}\:\frac{\alpha\:+\:\beta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\alpha\:−\:\beta}{\mathrm{2}}\:−\:\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\alpha\:+\:\beta}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\alpha\:+\:\beta}{\mathrm{2}}\:−\:\mathrm{4}\:\mathrm{cos}\:\frac{\alpha\:+\:\beta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\alpha\:−\:\beta}{\mathrm{2}}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}\:\mathrm{cos}\:\frac{\alpha\:+\:\beta}{\mathrm{2}}\:−\:\mathrm{cos}\:\frac{\alpha\:−\:\beta}{\mathrm{2}}\right)^{\mathrm{2}} \:+\:\mathrm{sin}^{\mathrm{2}} \:\frac{\alpha\:−\:\beta}{\mathrm{2}}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\alpha\:=\:\beta\:=\:\frac{\pi}{\mathrm{3}} \\ $$