Question

已知函数f（x）＝根号3sinxcosx＋cos²x

已知函数f（x）＝根号3sinxcosx＋cos²x求函数f（x）的最小正周期

Answer 1

f（x）＝根号3sinxcosx＋cos²x
=√3/2sin2x+(1+cos2x)/2
=√3/2sin2x+1/2cos2x+1/2
=sin(2x+π/6)+1/2
故T=2π/2=π。

Answer 2

f（x）＝根号3sinxcosx＋cos²x
= (√3/2)sin2x+(1/2)cos2x+1/2
= sin2xcosπ/6+cos2xsinπ/6+1/2
= sin(2x+π/6)
函数f（x）的最小正周期=2π/2 = π

Answer 3

f(x)=√3sinxcosx+cos²x
=(√3/2)·2sinxcosx+½(2cos²x-1)+½
=(√3/2)sin2x+½cos2x+½
=sin(2x+π/6) +½
最小正周期T=2π/2=π