作业帮已知函数f（x）=sin^2ωx+√3 sinωxsin(ωx+π/2)（ω＞0）的最小正周期为π（1） 求f(x)(2) 当x∈[-π/12,π/2]时,求函数f（x）的值域
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已知函数f（x）=sin^2ωx+√3 sinωxsin(ωx+π/2)（ω＞0）的最小正周期为π（1） 求f(x)(2) 当x∈[-π/12,π/2]时,求函数f（x）的值域
已知函数f（x）=sin^2ωx+√3 sinωxsin(ωx+π/2)（ω＞0）的最小正周期为π
（1） 求f(x)
(2) 当x∈[-π/12,π/2]时,求函数f（x）的值域
已知函数f（x）=sin^2ωx+√3 sinωxsin(ωx+π/2)（ω＞0）的最小正周期为π（1） 求f(x)(2) 当x∈[-π/12,π/2]时,求函数f（x）的值域
(1)
f(x) = sin²(ωx) + √3 sin(ωx)sin(ωx+π/2) = sin²(ωx) + √3sin(ωx)cos(ωx)
= 2[(1/2)sin(ωx) + (√3/2)cos(ωx)]sin(ωx)
= 2[cos(π/3)sin(ωx) + sin(π/3)cos(ωx)]sin(ωx)
= 2sin(ωx + π/3)sin(ωx)
= 2*(1/2)[cos(ωx + π/3 - ωx) - cos(ωx + π/3 + ωx)]
= -cos(2ωx + π/3) + cos(π/3)
= -cos(2ωx + π/3) + 1/2
= cos(2ωx + π/3 - π) + 1/2
= cos(2ωx - 2π/3) + 1/2
最小正周期为π = 2π/(2ω),ω = 1
f(x) = cos(2x - 2π/3) + 1/2
(2)
x∈[-π/12,π/2]:
f(π/3) = cos(2π/3 - 2π/3) + 1/2 = 1 + 1/2 = 3/2,此为最大值
x = π/3为f(x)最大值处的对称轴
π/3 - (-π/12) = 5π/12
π/2 - π/3 = 2π/12 < 5π/12
x = -π/12比x = π/2与x = π/3对称轴更远,最小值f(-π/12) = cos(-π/6 - 2π/3) + 1/2
= (1 - √3)/2
值域:[(1 - √3)/2,3/2]