Question Number 35886 by behi83417@gmail.com last updated on 25/May/18
Answered by ajfour last updated on 25/May/18
![sin (x+y)−sin x = sin y 2cos (x+(y/2))sin (y/2)=2cos (y/2)sin (y/2) ⇒ sin (y/2)[cos (x+(y/2))−cos (y/2)]=0 or (sin (y/2))(sin (x/2))sin (((x+y)/2))=0 ⇒ y=2mπ , or x=2nπ, or x+y=2kπ and since x−y=(π/3) we have x=2nπ, y=2nπ−(π/3) or y=2mπ, x=2mπ+(π/3) or x=kπ+(π/6), y=kπ−(π/6) .](https://www.tinkutara.com/question/Q35889.png)
$$\mathrm{sin}\:\left({x}+{y}\right)−\mathrm{sin}\:{x}\:=\:\mathrm{sin}\:{y} \\ $$$$\mathrm{2cos}\:\left({x}+\frac{{y}}{\mathrm{2}}\right)\mathrm{sin}\:\frac{{y}}{\mathrm{2}}=\mathrm{2cos}\:\frac{{y}}{\mathrm{2}}\mathrm{sin}\:\frac{{y}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{sin}\:\frac{{y}}{\mathrm{2}}\left[\mathrm{cos}\:\left({x}+\frac{{y}}{\mathrm{2}}\right)−\mathrm{cos}\:\frac{{y}}{\mathrm{2}}\right]=\mathrm{0} \\ $$$${or}\:\:\left(\mathrm{sin}\:\frac{{y}}{\mathrm{2}}\right)\left(\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{{x}+{y}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{y}=\mathrm{2}{m}\pi\:,\:{or}\:\:{x}=\mathrm{2}{n}\pi,\:{or} \\ $$$$\:\:\:\:\:\:\:\:\:{x}+{y}=\mathrm{2}{k}\pi \\ $$$${and}\:{since}\:\:{x}−{y}=\frac{\pi}{\mathrm{3}}\:\:{we}\:{have} \\ $$$$\:{x}=\mathrm{2}{n}\pi,\:{y}=\mathrm{2}{n}\pi−\frac{\pi}{\mathrm{3}} \\ $$$$\:{or}\:\:{y}=\mathrm{2}{m}\pi,\:\:{x}=\mathrm{2}{m}\pi+\frac{\pi}{\mathrm{3}} \\ $$$${or} \\ $$$$\:{x}={k}\pi+\frac{\pi}{\mathrm{6}},\:{y}={k}\pi−\frac{\pi}{\mathrm{6}}\:. \\ $$
Commented by behi83417@gmail.com last updated on 25/May/18
$${thank}\:{you}\:{very}\:{much}\:{sir}\:{Ajfour}. \\ $$