已知函数fx=cos(2x-派/3)-cos2x.①求函数fx的最小正周期.f(x)=cos(2x-π/3)-cos2x =1/2cos2x+√3/2sin2x-cos2x =√3/2sin2x-1/2cos2x(这一步是怎么转化的,我转化出来是=sin(2x-π/3)) =sin(2x-π/6)最小正周期T=2π/2=π
问题描述:
已知函数fx=cos(2x-派/3)-cos2x.①求函数fx的最小正周期.
f(x)=cos(2x-π/3)-cos2x
=1/2cos2x+√3/2sin2x-cos2x
=√3/2sin2x-1/2cos2x
(这一步是怎么转化的,我转化出来是=sin(2x-π/3))
=sin(2x-π/6)
最小正周期T=2π/2=π
答