Question

已知函数f(x)=2sin^2 ωx+2根号3sin ωx sin(π/2-ωx) ω＞0 的最小正周期为π

1 求ω的值 2 求函数f(x)在区间【0，2π/3】上的值域

Answer 1

1-cos2ωx+2√3sinωxcosωx
=√3sin2ωx-cos2ωx+1
=2sin(2ωx-π/6)+1

Answer 2

邋遢way聚聚估计

Answer 3

由于f(x)=2sin^2 ωx+2sqrt(3)sin ωx sin(π/2-ωx)
=2*(1-cos2wx)/2+2sqrt(3)sinwx*coswx
=1-cos2wx+sqrt(3)*sin2wx
=2[sqrt(3)/2*sin2wx-1/2cos2wx]+1
=2sin(2wx-π/6)+1
所以T=2π/2w=π/w=π
所以w=1.
(2) 由(1)知：f(x)=2sin(2x-π/6)+1
因为x属于[0,2π/3], 所以2x-π/6属于[-π/6, 7π/6]
从而sin(2x-π/6)属于[-1/2, 1]
因此f(x)=2sin(2x-π/6)+1的值域是[0,3]