Question

已知函数fx=cos^2x+2√3sinx*cosx-sin^2x则1.求fx的最小正周期和值域

1.求fx的最小正周期和值域

Answer 1

答：
f(x)=(cosx)^2+2√3sinxcosx-(sinx)^2
=cos2x+√3sin2x
=2*[(1/2)cos2x+(√3/2)sin2x]
=2sin(2x+π/6)
f(x)的最小正周期T=2π/2=π
值域为[-2,2]

Answer 2

解：由原式化为，f(x)=cos^2x-sin^2x+2√3sinx*cosx=cos2x+√3sin2x=2sin(2x+π/6)，
∴ f(x)的最小正周期为T=2π/2=π，值域为[-2,2].