Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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)

$${sin}\mathrm{2}{x}−{sin}\mathrm{2}{y}=−\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$\mathrm{2}{cos}\left({x}+{y}\right){sin}\left({x}−{y}\right)=−\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$−\frac{\mathrm{4}}{\mathrm{3}}{sin}^{\mathrm{2}} \left({x}−{y}\right)=−\frac{\mathrm{5}}{\mathrm{12}} \\ $$$${sin}\left({x}−{y}\right)=\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$${so}\:{cos}\left({x}+{y}\right)=\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{6}} \\ $$$${cos}\left({x}+{y}\right)+\mathrm{2}{sin}\left({x}−{y}\right)=\pm\left(−\frac{\sqrt{\mathrm{5}}}{\mathrm{6}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)=\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com