Question

函数f(x)=cos2x-2根号3sinxcosx的最小正周期是？

Answer 1

f(x)=cos2x-2根号3sinxcosx=cos2x-根号3sin2x =2（cos2x/2-根号3sin2x/2）=2（cos2xcos60度-sin2xsin60度）=2cos（2x+60度）最小正周期：2pai/2=pai就是pai

Answer 2

f(x)=cos2x-2√3sinxcosx=cos2x-√3sin2x=2*[(1/2)*cos2x - (√3 /2)sin2x]=2[cos(π/3)cos2x - sin(π/3)sin2x]=2cos(2x + π/3)所以最小正周期=2π/2=π

Answer 3

f(x)=cos2x-√(3)sin2x
=2（1/2cos2x-√3/2sin2x)
=2(cos2xsinπ/6-sin2xcosπ/6)
=2sin(2x-π/6)
所以最小正周期T=2π/2=π