Question

已知函数f(x)=√3sinxcosx-cos²x-cos²+1/2(x∈R)

(1)求函数f(x)的最小正周期
（2）求f(x)的单调区间
（3）求f(x)在区间（0，π、3】上的函数值的取值范围

Answer 1

∵ cos2x=2cos²x-1
∴ cos²x=(cos2x+1)/2
∵ sin2x=2sinxcosx
∴ f(x)=√3sinxcosx-cos²x+1/2
=√3/2sin2x-1/2cos2x-1/2+1/2
=√3/2sin2x-1/2cos2x
=sin2xcosπ/6-sinπ/6cos2x
=sin(2x-π/6)
(1) 函数f(x)的最小正周期为：2π/2=π

(2) 单调增：2kπ-π/2<2x-π/6≤π/2+2kπ
2kπ-π/3<2x≤2π/3+2kπ
即 kπ-π/6<x≤π/3+kπ
单调减：2kπ+π/2<2x-π/6≤3π/2+2kπ
2kπ+2π/3<2x≤5π/3+2kπ
即 kπ+π/3<x≤5π/6+kπ
(3) f(x)在区间(0，π/3]的取值范围
f(x)∈(-1/2，1]