Question

已知函数fx=cosx·sin(x+π/3)-√3cos^2x+√3/4x∈r求fx的最小正周期

Answer 1

f(x)=cosx·sin(x+π/3)-√3cos²x+√3/4
=sinxcosxcosπ/3+cos²xsinπ/3-√3cos²x+√3/4
=1/4sin2x+√3/4cos2x
=1/2[sin2xcosπ/3+sinπ/3cos2x)]
=1/2sin(2x+π/3)
f(x)的最小正周期：T=2π/2=π

Answer 2

f(x)=cosx·sin(x+π/3)-√3cos^2x+√3/4
=1/2cosx·sinx+√3/2cos^2x-√3cos^2x+√3/4
=1/4sin2x-√3/2cos^2x+√3/4
=1/4sin2x-√3/4-√3/4cos2x+√3/4
=1/4sin2x-√3/4cos2x
=1/2sin(2x-π/6)
T=2π/2=π

追问

求fx在［-π/4，π/4］上的最大值和最小值

追答

x在［-π/4，π/4］

2x-π/6在［-2π/3，π/3］
最大值=√3/4

和最小值=-1/2