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Question Number 105861 by bemath last updated on 01/Aug/20
Given { ((sin 2x−sin 2y=−(5/(12)))),((cos (x+y)= −(2/3)sin (x−y))) :} where 0 < x−y<π. find the value of cos (x+y)+2sin (x−y)
$$\mathcal{G}{iven}\:\begin{cases}{\mathrm{sin}\:\mathrm{2}{x}−\mathrm{sin}\:\mathrm{2}{y}=−\frac{\mathrm{5}}{\mathrm{12}}}\\{\mathrm{cos}\:\left({x}+{y}\right)=\:−\frac{\mathrm{2}}{\mathrm{3}}\mathrm{sin}\:\left({x}−{y}\right)}\end{cases} \\ $$$${where}\:\mathrm{0}\:<\:{x}−{y}<\pi. \\ $$$${find}\:{the}\:{value}\:{of}\:\mathrm{cos}\:\left({x}+{y}\right)+\mathrm{2sin}\:\left({x}−{y}\right) \\ $$
Commented by PRITHWISH SEN 2 last updated on 01/Aug/20
sin 2x−sin 2y=−(5/(12)) ⇒24sin (x−y)cos (x+y)=−5 ......(i) 3cos (x+y)+2sin (x−y)=0.......(ii) 3cos (x+y)−2sin (x−y)= (√({3cos(x+y)+2sin (x−y)}^2 −24sin (x−y)cos (x+y) )) = ±(√5) .....(iii) from (ii) − (iii) 4sin (x−y)=±(√5) Again given cos (x+y)=−(2/3)sin (x−y) ⇒((cos (x+y))/(2sin (x−y))) +1 = 1−(1/3) ⇒ cos (x+y)+2sin (x−y)=((4sin (x−y))/3) = ±((√5)/3)
$$\mathrm{sin}\:\mathrm{2x}−\mathrm{sin}\:\mathrm{2y}=−\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$\Rightarrow\mathrm{24sin}\:\left(\mathrm{x}−\mathrm{y}\right)\mathrm{cos}\:\left(\mathrm{x}+\mathrm{y}\right)=−\mathrm{5}\:……\left(\mathrm{i}\right) \\ $$$$\mathrm{3cos}\:\left(\mathrm{x}+\mathrm{y}\right)+\mathrm{2sin}\:\left(\mathrm{x}−\mathrm{y}\right)=\mathrm{0}…….\left(\mathrm{ii}\right) \\ $$$$\mathrm{3cos}\:\left(\mathrm{x}+\mathrm{y}\right)−\mathrm{2sin}\:\left(\mathrm{x}−\mathrm{y}\right)=\:\sqrt{\left\{\mathrm{3cos}\left(\mathrm{x}+\mathrm{y}\right)+\mathrm{2sin}\:\left(\mathrm{x}−\mathrm{y}\right)\right\}^{\mathrm{2}} −\mathrm{24sin}\:\left(\mathrm{x}−\mathrm{y}\right)\mathrm{cos}\:\left(\mathrm{x}+\mathrm{y}\right)\:} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\pm\sqrt{\mathrm{5}}\:…..\left(\mathrm{iii}\right) \\ $$$$\mathrm{from}\:\left(\mathrm{ii}\right)\:−\:\left(\mathrm{iii}\right) \\ $$$$\mathrm{4sin}\:\left(\mathrm{x}−\mathrm{y}\right)=\pm\sqrt{\mathrm{5}} \\ $$$$\mathrm{Again}\:\mathrm{given} \\ $$$$\mathrm{cos}\:\left(\mathrm{x}+\mathrm{y}\right)=−\frac{\mathrm{2}}{\mathrm{3}}\mathrm{sin}\:\left(\mathrm{x}−\mathrm{y}\right) \\ $$$$\Rightarrow\frac{\mathrm{cos}\:\left(\mathrm{x}+\mathrm{y}\right)}{\mathrm{2sin}\:\left(\mathrm{x}−\mathrm{y}\right)}\:+\mathrm{1}\:=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{cos}\:\left(\mathrm{x}+\mathrm{y}\right)+\mathrm{2sin}\:\left(\mathrm{x}−\mathrm{y}\right)=\frac{\mathrm{4sin}\:\left(\mathrm{x}−\mathrm{y}\right)}{\mathrm{3}}\:=\:\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$
Answered by Dwaipayan Shikari last updated on 01/Aug/20
sin2x−sin2y=−(5/(12)) 2cos(x+y)sin(x−y)=−(5/(12)) −(4/3)sin^2 (x−y)=−(5/(12)) sin(x−y)=±((√5)/4) so cos(x+y)=±((√5)/6) cos(x+y)+2sin(x−y)=±(−((√5)/6)+((√5)/2))=±((√5)/3)
$${sin}\mathrm{2}{x}−{sin}\mathrm{2}{y}=−\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$\mathrm{2}{cos}\left({x}+{y}\right){sin}\left({x}−{y}\right)=−\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$−\frac{\mathrm{4}}{\mathrm{3}}{sin}^{\mathrm{2}} \left({x}−{y}\right)=−\frac{\mathrm{5}}{\mathrm{12}} \\ $$$${sin}\left({x}−{y}\right)=\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$${so}\:{cos}\left({x}+{y}\right)=\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{6}} \\ $$$${cos}\left({x}+{y}\right)+\mathrm{2}{sin}\left({x}−{y}\right)=\pm\left(−\frac{\sqrt{\mathrm{5}}}{\mathrm{6}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)=\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$

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