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Question Number 105243 by bemath last updated on 27/Jul/20

$$\{\begin{array}{l}\sin 2x+\sin 2y=\frac{4}{9}\\ \cos (x-y)=1-\sin (x+y)\end{array}$$ $$0<x<\frac{\pi }{2};0<y<\frac{\pi }{2}$$ $$findthevalueof\sin (x+y)$$ $$\left(a\right)-\frac{2}{3}\left(b\right)-\frac{1}{3}\left(c\right)\frac{1}{9}$$ $$\left(d\right)\frac{2}{9}\left(e\right)\frac{2}{3}$$

Answered by bramlex last updated on 27/Jul/20

$$\Rightarrow 2\sin (x+y)\cos (x-y)=\frac{4}{9}...\left(1\right)$$ $$\Rightarrow \cos (x-y)=1-\sin (x+y)...\left(2\right)$$ $$let\sin (x+y)=v$$ $$\Rightarrow 2v.(1-v)=\frac{4}{9}$$ $$2v^2-2v+\frac{4}{9}=0$$ $$v^2-v+\frac{2}{9}=0$$ $$v=\frac{1+\sqrt{1-4.\left(\frac{2}{9}\right)}}{2}=\frac{1+\frac{1}{3}}{2}$$ $$\therefore \sin (x+y)=\frac{4}{6}=\frac{2}{3}$$ $$note0<x+y<\pi then\sin (x+y)>0$$

Commented bybemath last updated on 27/Jul/20

$$afdolll$$

Answered by Dwaipayan Shikari last updated on 27/Jul/20

$$sin2x+sin2y$$ $$=2sin(x+y)cos(x-y)$$ $$\Rightarrow 2sin(x+y)(1-sin(x+y))=\frac{4}{9}$$ $$\Rightarrow 2{sin}^2(x+y)-2sin(x+y)+\frac{4}{9}=0$$ $$\Rightarrow sin(x+y)=\frac{2+\sqrt{4-\frac{32}{9}}}{4}=\frac{2+\frac{2}{3}}{4}=\frac{2}{3}$$

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