Question

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

（1）求函数f（x）的单调递增区间
（2）求函数f（x）在区间【0，派／3】上的取值范围

Answer 1

解：
(1) f（x）=sin²ωx+√3sinωx sin(ωx+π/2)
=sin²ωx+√3sinωx cosωx
= (1/2)-(1/2)cos2ωx + (√3/2)sin 2ωx
= (1/2) + (√3/2)sin 2ωx - (1/2)cos2ωx
= (1/2) + sin 2ωx cosπ/6 - cos2ωx sinπ/6
= (1/2) + sin(2ωx - π/6)
其最小正周期为 T = 2π/2ω = π/ω =π/2
∴ ω =2
故 f（x）= sin(4x - π/6) + 1/2
单调递增区间 由双向不等式 2kπ -π/2 ≤ 4x - π/6 ≤ 2kπ + π/2 解出：
2kπ -π/2 ≤ 4x - π/6 ≤ 2kπ +π/2
→ 2kπ -π/3 ≤ 4x ≤ 2kπ -2π/3
→ kπ/2 - π/12 ≤ x ≤ kπ/2 - π/6

(2) 0 ≤ x ≤ π/3
→ 0 ≤ 4x ≤ 4π/3
→ 0 - π/6 ≤ 4x - π/6 ≤ 4π/3 - π/6
→ - π/6 ≤ 4x - π/6 ≤ 7π/6
→ sin(- π/6) ≤ sin(4x - π/6) ≤ sin(7π/6)
→ -1/2 ≤ sin(4x - π/6) ≤ 1
→ 0 ≤ sin(4x - π/6)+1/2 ≤ 3/2
即 0 ≤ f（x） ≤ 3/2