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
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(2π
1个回答
相关问题
-
看不懂这一步?y=sin(x-π/3)cosx=1/2[sin(2x-π/3)+sin(-π/3)]
-
y=sin(π/4+x/2)sin(π/4-x/2) =sin(π/4+x/2)sin[π/2-(π/4+x/2)]
-
y=2sin三分之一x y=4sin(x-π/3) y=sin(2x+π/6) y=3sin(1/2x-π/4) 的周期
-
若M={(x,y)||tanπy|+sin^2πx=0},N={(x,y)|x^2+y^2
-
求值域 (1)y=2sin(x/2+π/4)x∈[0,π) (2)y=2cosx-sin^2x+3 x∈[-π/3,2/
-
y=2sin(x/2+π/6),x属于R y=-6sin(2x-π/3),x属于R 写出振幅,周期,初相
-
函数y=sin(x+π2),x∈R是( )
-
函数y=sin(x+π2),x∈R是( )
-
y=3sin(2x+π/5)怎么变化得到y=2sin(2x+π/4)
-
y=sin(x+π/6)+sin(x-π/6)+cosx ,x∈[-π/2,π/2]的值域