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Question Number 105243 by bemath last updated on 27/Jul/20
 { ((sin 2x + sin 2y = (4/9))),((cos (x−y) = 1−sin (x+y))) :}  0 < x < (π/2) ; 0 < y < (π/2)  find the value of sin (x+y)  (a) −(2/3)   (b) −(1/3)   (c) (1/9)  (d) (2/9)    (e) (2/3)
$$\begin{cases}{\mathrm{sin}\:\mathrm{2}{x}\:+\:\mathrm{sin}\:\mathrm{2}{y}\:=\:\frac{\mathrm{4}}{\mathrm{9}}}\\{\mathrm{cos}\:\left({x}−{y}\right)\:=\:\mathrm{1}−\mathrm{sin}\:\left({x}+{y}\right)}\end{cases} \\ $$$$\mathrm{0}\:<\:{x}\:<\:\frac{\pi}{\mathrm{2}}\:;\:\mathrm{0}\:<\:{y}\:<\:\frac{\pi}{\mathrm{2}} \\ $$$${find}\:{the}\:{value}\:{of}\:\mathrm{sin}\:\left({x}+{y}\right) \\ $$$$\left({a}\right)\:−\frac{\mathrm{2}}{\mathrm{3}}\:\:\:\left({b}\right)\:−\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\left({c}\right)\:\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\left({d}\right)\:\frac{\mathrm{2}}{\mathrm{9}}\:\:\:\:\left({e}\right)\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Answered by bramlex last updated on 27/Jul/20
⇒ 2sin (x+y)cos (x−y)=(4/9)...(1)  ⇒cos (x−y) = 1−sin(x+y)...(2)  let sin (x+y) = v  ⇒2v.(1−v) = (4/9)  2v^2 −2v + (4/9) = 0  v^2 −v + (2/9) = 0  v = ((1 + (√(1−4.((2/9)))))/2) = ((1+(1/3))/2)  ∴ sin (x+y) = (4/6) = (2/3)  note 0 < x+y < π then sin(x+y)>0
$$\Rightarrow\:\mathrm{2sin}\:\left({x}+{y}\right)\mathrm{cos}\:\left({x}−{y}\right)=\frac{\mathrm{4}}{\mathrm{9}}…\left(\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{cos}\:\left({x}−{y}\right)\:=\:\mathrm{1}−\mathrm{sin}\left({x}+{y}\right)…\left(\mathrm{2}\right) \\ $$$${let}\:\mathrm{sin}\:\left({x}+{y}\right)\:=\:{v} \\ $$$$\Rightarrow\mathrm{2}{v}.\left(\mathrm{1}−{v}\right)\:=\:\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\mathrm{2}{v}^{\mathrm{2}} −\mathrm{2}{v}\:+\:\frac{\mathrm{4}}{\mathrm{9}}\:=\:\mathrm{0} \\ $$$${v}^{\mathrm{2}} −{v}\:+\:\frac{\mathrm{2}}{\mathrm{9}}\:=\:\mathrm{0} \\ $$$${v}\:=\:\frac{\mathrm{1}\:+\:\sqrt{\mathrm{1}−\mathrm{4}.\left(\frac{\mathrm{2}}{\mathrm{9}}\right)}}{\mathrm{2}}\:=\:\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{2}} \\ $$$$\therefore\:\mathrm{sin}\:\left({x}+{y}\right)\:=\:\frac{\mathrm{4}}{\mathrm{6}}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${note}\:\mathrm{0}\:<\:{x}+{y}\:<\:\pi\:{then}\:\mathrm{sin}\left({x}+{y}\right)>\mathrm{0} \\ $$
Commented by bemath last updated on 27/Jul/20
afdolll
$${afdolll} \\ $$
Answered by Dwaipayan Shikari last updated on 27/Jul/20
sin2x+sin2y  =2sin(x+y)cos(x−y)  ⇒2sin(x+y)(1−sin(x+y))=(4/9)  ⇒2sin^2 (x+y)−2sin(x+y)+(4/9)=0  ⇒sin(x+y)=((2+(√(4−((32)/9))))/4)=((2+(2/3))/4)=(2/3)
$${sin}\mathrm{2}{x}+{sin}\mathrm{2}{y} \\ $$$$=\mathrm{2}{sin}\left({x}+{y}\right){cos}\left({x}−{y}\right) \\ $$$$\Rightarrow\mathrm{2}{sin}\left({x}+{y}\right)\left(\mathrm{1}−{sin}\left({x}+{y}\right)\right)=\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\Rightarrow\mathrm{2}{sin}^{\mathrm{2}} \left({x}+{y}\right)−\mathrm{2}{sin}\left({x}+{y}\right)+\frac{\mathrm{4}}{\mathrm{9}}=\mathrm{0} \\ $$$$\Rightarrow{sin}\left({x}+{y}\right)=\frac{\mathrm{2}+\sqrt{\mathrm{4}−\frac{\mathrm{32}}{\mathrm{9}}}}{\mathrm{4}}=\frac{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{4}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$

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