Question

已知函数f(x)=sin^2wx+√3sinwxsin(wx+π/2)(w>0)的最小正周期为π

Answer 1

f(x)=(1-cos2wx)/2+√3/2*sin2x

=(√3/2)sin2wx-1/2*cos2wx+1/2

=√[(√3/2)^2+(1/2)^2]*sin(2wx-z)+1/2

其中tanz=(1/2)/(√3/2)

所以z=π/6

所以f(x)=sin(2wx-π/6/6)+1/2

T=2π/|2w|=π

w>0

所以w=1

f(x)=sin(2x-π/6)+1/2

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