Question

f(x)=sin(x+7π\4)+cos(x-3π\4) 求fx的最小正周期

要化简过程

Answer 1

f(x)=sin(x+7π\4)+cos(x-3π\4)
=sinxcos7π/4+cosxsin7π/4+cosxcos3π/4+sinxsin3π/4
=sinxcosπ/4-cosxsinπ/4-cosxcosπ/4+sinxsinπ/4
=根号2/2sinx-根号2/2cosx-根号2/2cosx+根号2/2sinx
=根号2sinx-根号2cosx
=2(根号2/2sinx-根号2/2cosx)
=2sin(x-π/4)
T=2π/1=2π

Answer 2

f(x)=sin(x+7π/4)+cos(x-3π/4)
=（1/√2）（sinx-cosx-cosx+sinx)
=2sin(x- π/4),
它的最小正周期是2π。

Answer 3

f(x)=sin(x+7π\4)+cos(x-3π\4)
=sinxcos7π/4+cosxsin7π/4+cosxcos3π/4+sinxsin3π/4
=sinxcosπ/4-cosxsinπ/4-cosxcosπ/4+sinxsinπ/4
=（1/√2）（sinx-cosx-cosx+sinx)
=2sin(x- π/4),
它的最小正周期是2π。

Answer 4

2排