Question

求下列函数的最大值,最小值和最小正周期y=sin(x+π/4)+sin(x-π/4)

具体过程

Answer 1

sin(x+π/4)+sin(x-π/4)=sin(x+π/4)+sin[(x+π/4)-π/2]=sin(x+π/4)-sin(x+π/4)cos(π/2)-cos(x+π/4)sin(π/2)=sin(x+π/4)-cos(x+π/4)=根号2[根号2/2sin(x+π/4)-根号2/2cos(x+π/4）]=根号2[sin(x+π/4)cos（π/4)-cos(x+π/4）sin(π/4)]=根号2sin(x+π/4-π/4)=根号2sin(x)
最大值根号2 最小值 -根号2 最小正周期2π

Answer 2

y=sin(x+π/4)+sin(x-π/4)
=sinxcosπ/4+cosxsinπ/4+sinxcosπ/4-cosxsinπ/4
=2sinxcosπ/4
=√2sinx
最大值√2
最小值-√2
最小正周期2π

Answer 3

化简 y=√2sinx 所以y max=√2 ymin=-√2 最小正周期 2π

Answer 4

解:y=sin(x＋派/4)＋sin(x-派/4)=根号2sinx,所以函数y的最大值为根号2,最小值为-根号2,函数周期T=2派