f(x)=-√3*sin²x+sinxcosx
=-√3*(1-cos2x)/2+1/2sin2x
=1/2sin2x+√3/2cos2x-√3/2
= sin(2x+π/3) -√3/2
最小正周期是π.
x∈[0,π／2]時
2x+π/3∈[π/3,4π/3],
sin(2x+π/3) ∈[-√3/2,1]
所以sin(2x+π/3) -√3/2∈[-√3,1-√3/2].
函數(shù)值域是[-√3,1-√3/2].