Question Number 193595 by Rupesh123 last updated on 17/Jun/23
Answered by MM42 last updated on 17/Jun/23
$${sin}^{\mathrm{2}} \alpha+{sin}^{\mathrm{2}} \beta−{sin}\alpha{cos}\beta−{sim}\beta{cos}\alpha=\mathrm{0} \\ $$$$\Rightarrow{sin}\alpha\left({sin}\alpha−{cos}\beta\right)+{sin}\beta\left({sin}\beta−{cos}\alpha\right)=\mathrm{0} \\ $$$${sin}\alpha\left(\mathrm{2}{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right){cos}\left(\frac{\alpha−\beta}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)+{sin}\beta\left(\mathrm{2}{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right){cos}\left(\frac{\beta−\alpha}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)\left({sin}\alpha{cos}\left(\frac{\alpha−\beta}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)+{sin}\beta{cos}\left(\frac{\beta−\alpha}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)=\mathrm{0} \\ $$$${sin}\alpha{cos}\left(\frac{\alpha−\beta}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)+{sin}\beta{cos}\left(\frac{\beta−\alpha}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\neq\:\mathrm{0} \\ $$$$\Rightarrow{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)=\mathrm{0} \\ $$$$\Rightarrow\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}=\mathrm{0}\Rightarrow\:\alpha+\beta=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$
Commented by Rupesh123 last updated on 17/Jun/23
Perfect
Answered by Subhi last updated on 17/Jun/23
$${sin}\left(\alpha+\beta\right)={sin}\left(\alpha\right){cos}\left(\beta\right)+{cos}\left(\alpha\right){sin}\left(\beta\right)={sin}^{\mathrm{2}} \left(\alpha\right)+{sin}^{\mathrm{2}} \left(\beta\right) \\ $$$$\therefore\:{sin}\left(\alpha\right)\:=\:{cos}\left(\beta\right) \\ $$$${cos}\left(\alpha\right)\:=\:{sin}\left(\beta\right) \\ $$$${sin}\left(\alpha\right){cos}\left(\alpha\right)={sin}\left(\beta\right){cos}\left(\beta\right) \\ $$$$\frac{{sin}\left(\mathrm{2}\alpha\right)}{\mathrm{2}}=\frac{{sin}\left(\mathrm{2}\beta\right)}{\mathrm{2}}\:\Rrightarrow\:{sin}\left(\mathrm{2}\alpha\right)−{sin}\left(\mathrm{2}\beta\right)=\mathrm{0} \\ $$$$\mathrm{2}{sin}\left(\alpha−\beta\right){cos}\left(\alpha+\beta\right)=\mathrm{0} \\ $$$${cos}\left(\alpha+\beta\right)=\mathrm{0}\:\Rrightarrow\:\alpha+\beta=\mathrm{90}=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$
Commented by Rupesh123 last updated on 17/Jun/23
Perfect