第一步推论看不懂设S={(x,y)|x2－y2=奇数,x,y∈R},T={(x,y)|sin(2πx2)－sin(2πy2)=cos(2πx2)－cos(2πy2),x,y∈R},则()A．S((TB．T((SC．S=TD．S∩T若x2－y2为奇数,则sin(2πx2)－sin(2πy2)=cos(2πx2)－cos(2πy2)成立,即

Question

第一步推论看不懂
设S={(x,y)|x2－y2=奇数,x,y∈R},T={(x,y)|sin(2πx2)－sin(2πy2)=cos(2πx2)－cos(2πy2),x,y∈R},则( ) A．S((T B．T((S C．S=T D．S∩T 若x2－y2为奇数,则sin(2πx2)－sin(2πy2)=cos(2πx2)－cos(2πy2)成立,即S?T．　　又若x=y时,sin(2πx2)－sin(2πy2)=cos(2πx2)－cos(2πy2)也成立,即得S((T,选A

Answer 1

x2－y2为奇数,可设x2=y2+2k+1,带入方程即可验证成立,所以S是T的子集