Question

已知f(x)=(sinωx)^2+√3/2sin2ωx-1/2(x∈R,ω>0),若f(x)的最小正周期为2π.求f(x

Answer 1

f(x)=(sinωx)^2+√3/2sin2ωx-1/2

=1/2(1-cos2ωx)+√3/2sin2ωx-1/2

=√3/2sin2ωx-1/2cos2ωx

=sin(2ωx-π/6)

因为最小正周期是2π 所以2ω=1

所以f(x)=sin(x-π/6)