f(x)=cos(2x-2派/3)+1-2cos^2x,求函数f(x)的最小正周期及单调递增区间.

问题描述:

f(x)=cos(2x-2派/3)+1-2cos^2x,求函数f(x)的最小正周期及单调递增区间.

y=cos2xcos(2π/3)+sin2xsin(2π/3)-cos2x.=-(1/2)cos2x+(√3/2)sin2x-cos2x.=-(3/2)cos2x+(√3/2)sin2x.=√3[(1/2)sin2x-(√3/2)cos2x].=√3sin(2x-π/3)最小正周期是2π/2=π增区间是:2kπ-π/2≤2x-...