设函数f(x)=cos(2x+π/3)+sin2X,求最小正周期

设函数f(x)=cos(2x+π/3)+sin2X,求最小正周期
设函数f(x)=cos(2x+π/3)+sin2X,求最小正周期

设函数f(x)=cos(2x+π/3)+sin2X,求最小正周期
f(x)=cos(2x+π/3)+sin2x
=cos(2x+π/3)-cos(2x+π/2)
=2cos{[(2x+π/3)+(2x+π/2)]/2}cos{[(2x+π/3)-(2x+π/2)]/2}
=2cos(2x+5π/12)cos(-π/12)
=[(√6+√2)/2]cos(2x+5π/12)
频率ω=2,所以最小正周期为2π/2=π