Question Number 91038 by M±th+et+s last updated on 27/Apr/20
$${if}\:{sin}\left(\frac{\alpha}{\mathrm{2}}\right)=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${and}\:{cos}\left(\frac{\beta}{\mathrm{2}}\right)=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${prove} \\ $$$${sin}\left(\alpha\right)={cos}\left(\beta\right) \\ $$
Commented by jagoll last updated on 27/Apr/20
$$\mathrm{sin}\:\alpha\:=\:\mathrm{2sin}\:\left(\frac{\alpha}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\alpha}{\mathrm{2}}\right) \\ $$$$=\:\mathrm{2}×\frac{\mathrm{4}}{\mathrm{5}}×\frac{\mathrm{3}}{\mathrm{5}}\:=\:\frac{\mathrm{24}}{\mathrm{25}} \\ $$$$\mathrm{cos}\:\beta\:=\:\mathrm{2sin}\:\left(\frac{\beta}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\beta}{\mathrm{2}}\right) \\ $$$$=\:\mathrm{2}×\frac{\mathrm{3}}{\mathrm{5}}×\frac{\mathrm{4}}{\mathrm{5}}=\:\frac{\mathrm{24}}{\mathrm{25}} \\ $$
Answered by $@ty@m123 last updated on 27/Apr/20
$$\mathrm{sin}^{\mathrm{2}} \frac{\alpha}{\mathrm{2}}+\mathrm{cos}\:^{\mathrm{2}} \frac{\beta}{\mathrm{2}}=\frac{\mathrm{4}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }{\mathrm{5}^{\mathrm{2}} } \\ $$$$\mathrm{sin}^{\mathrm{2}} \frac{\alpha}{\mathrm{2}}+\mathrm{cos}\:^{\mathrm{2}} \frac{\beta}{\mathrm{2}}=\mathrm{1} \\ $$$$\Rightarrow\frac{\alpha}{\mathrm{2}}=\frac{\beta}{\mathrm{2}} \\ $$$$\Rightarrow\alpha=\beta \\ $$