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Question-193595

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Question Number 193595 by Rupesh123 last updated on 17/Jun/23
Answered by MM42 last updated on 17/Jun/23
sin^2 α+sin^2 β−sinαcosβ−simβcosα=0 ⇒sinα(sinα−cosβ)+sinβ(sinβ−cosα)=0 sinα(2sin(((α+β)/2)−(π/4))cos(((α−β)/2)+(π/4)))+sinβ(2sin(((α+β)/2)−(π/4))cos(((β−α)/2)+(π/4)))=0 ⇒sin(((α+β)/2)−(π/4))(sinαcos(((α−β)/2)+(π/4))+sinβcos(((β−α)/2)+(π/4)))=0 sinαcos(((α−β)/2)+(π/4))+sinβcos(((β−α)/2)+(π/4))≠ 0 ⇒sin(((α+β)/2)−(π/4))=0 ⇒((α+β)/2)−(π/4)=0⇒ α+β=(π/2)
$${sin}^{\mathrm{2}} \alpha+{sin}^{\mathrm{2}} \beta−{sin}\alpha{cos}\beta−{sim}\beta{cos}\alpha=\mathrm{0} \\ $$$$\Rightarrow{sin}\alpha\left({sin}\alpha−{cos}\beta\right)+{sin}\beta\left({sin}\beta−{cos}\alpha\right)=\mathrm{0} \\ $$$${sin}\alpha\left(\mathrm{2}{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right){cos}\left(\frac{\alpha−\beta}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)+{sin}\beta\left(\mathrm{2}{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right){cos}\left(\frac{\beta−\alpha}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)\left({sin}\alpha{cos}\left(\frac{\alpha−\beta}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)+{sin}\beta{cos}\left(\frac{\beta−\alpha}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)=\mathrm{0} \\ $$$${sin}\alpha{cos}\left(\frac{\alpha−\beta}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)+{sin}\beta{cos}\left(\frac{\beta−\alpha}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\neq\:\mathrm{0} \\ $$$$\Rightarrow{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)=\mathrm{0} \\ $$$$\Rightarrow\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}=\mathrm{0}\Rightarrow\:\alpha+\beta=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$
Commented by Rupesh123 last updated on 17/Jun/23
Perfect ��
Answered by Subhi last updated on 17/Jun/23
sin(α+β)=sin(α)cos(β)+cos(α)sin(β)=sin^2 (α)+sin^2 (β) ∴ sin(α) = cos(β) cos(α) = sin(β) sin(α)cos(α)=sin(β)cos(β) ((sin(2α))/2)=((sin(2β))/2) ⇛ sin(2α)−sin(2β)=0 2sin(α−β)cos(α+β)=0 cos(α+β)=0 ⇛ α+β=90=(π/2)
$${sin}\left(\alpha+\beta\right)={sin}\left(\alpha\right){cos}\left(\beta\right)+{cos}\left(\alpha\right){sin}\left(\beta\right)={sin}^{\mathrm{2}} \left(\alpha\right)+{sin}^{\mathrm{2}} \left(\beta\right) \\ $$$$\therefore\:{sin}\left(\alpha\right)\:=\:{cos}\left(\beta\right) \\ $$$${cos}\left(\alpha\right)\:=\:{sin}\left(\beta\right) \\ $$$${sin}\left(\alpha\right){cos}\left(\alpha\right)={sin}\left(\beta\right){cos}\left(\beta\right) \\ $$$$\frac{{sin}\left(\mathrm{2}\alpha\right)}{\mathrm{2}}=\frac{{sin}\left(\mathrm{2}\beta\right)}{\mathrm{2}}\:\Rrightarrow\:{sin}\left(\mathrm{2}\alpha\right)−{sin}\left(\mathrm{2}\beta\right)=\mathrm{0} \\ $$$$\mathrm{2}{sin}\left(\alpha−\beta\right){cos}\left(\alpha+\beta\right)=\mathrm{0} \\ $$$${cos}\left(\alpha+\beta\right)=\mathrm{0}\:\Rrightarrow\:\alpha+\beta=\mathrm{90}=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$
Commented by Rupesh123 last updated on 17/Jun/23
Perfect ��

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