Menu Close

Question-35886




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) .
$$\mathrm{sin}\:\left({x}+{y}\right)−\mathrm{sin}\:{x}\:=\:\mathrm{sin}\:{y} \\ $$$$\mathrm{2cos}\:\left({x}+\frac{{y}}{\mathrm{2}}\right)\mathrm{sin}\:\frac{{y}}{\mathrm{2}}=\mathrm{2cos}\:\frac{{y}}{\mathrm{2}}\mathrm{sin}\:\frac{{y}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{sin}\:\frac{{y}}{\mathrm{2}}\left[\mathrm{cos}\:\left({x}+\frac{{y}}{\mathrm{2}}\right)−\mathrm{cos}\:\frac{{y}}{\mathrm{2}}\right]=\mathrm{0} \\ $$$${or}\:\:\left(\mathrm{sin}\:\frac{{y}}{\mathrm{2}}\right)\left(\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{{x}+{y}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{y}=\mathrm{2}{m}\pi\:,\:{or}\:\:{x}=\mathrm{2}{n}\pi,\:{or} \\ $$$$\:\:\:\:\:\:\:\:\:{x}+{y}=\mathrm{2}{k}\pi \\ $$$${and}\:{since}\:\:{x}−{y}=\frac{\pi}{\mathrm{3}}\:\:{we}\:{have} \\ $$$$\:{x}=\mathrm{2}{n}\pi,\:{y}=\mathrm{2}{n}\pi−\frac{\pi}{\mathrm{3}} \\ $$$$\:{or}\:\:{y}=\mathrm{2}{m}\pi,\:\:{x}=\mathrm{2}{m}\pi+\frac{\pi}{\mathrm{3}} \\ $$$${or} \\ $$$$\:{x}={k}\pi+\frac{\pi}{\mathrm{6}},\:{y}={k}\pi−\frac{\pi}{\mathrm{6}}\:. \\ $$
Commented by behi83417@gmail.com last updated on 25/May/18
thank you very much sir Ajfour.
$${thank}\:{you}\:{very}\:{much}\:{sir}\:{Ajfour}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *