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

解：因为函数f(x)=sin^2ωx+√3sinωxsin(ωx+π/2),(ω>0)=根号3sin2wx/2-con2wx/2+1/2 =sin(2wx-π/6)+1/2 因为最小正周期为π所以w=1 2x-6/π属于[-13π/6,17π/6]所以f(x）在［-π，3π／2］上的单调区间是 递增区间是[-pi,-2pi/3]还有其他的部分，可自行...

(1)解析：∵函数f(x)=sin^2wx+√3sinwxsin(π/2+wx) (w>0) 的最小正周期为π f(x)=sin^2wx+√3sinwxsin(π/2+wx)=1/2-1/2cos2wx+√3/2sin2wx =sin(2wx-π/6)+1/2 ∵T=π==>2w=2==>w=1 ∴f(x)=sin(2x-π/6)+1/2 ∴单调递增区间：2kπ-π/2<=2x-π...

= 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 - (-...

最小正周期=2π/2w=π w=1 2)在区间[0,2π/3]上 2x-π/6=π/2,x=π/3时，f(x)有最大值=1/2+1=3/2 x=0时，f(x)有最小值=1/2-1/2=0 函数f(x)在区间[0,2π/3]上的取值范围 :[0,3/2]

解：f(x)=（1-cos2ωx)/2+√3sinωxcosωx.=(√3/2)sin2ωx-(1/2)cos2ωx+1/2.=sin(2ωx-π/6)+1/2.∵T=2π/2ω=π.∴ω=1.∴f(x)=sin(2x-π/6)+1/2.当sin(2x-π/6)=1, x=5π/12时，sin(2x-π/6)取得最大值1，函数f(x)取得最大值，f(x)max=1+...

1. f(x)=sin²ωx+根号3sinωxsin[ωx+π/2]=1/2-(1/2)cos2wx+√3sinwxcoswx =(√3/2)sin2wx-(1/2)cos2wx+1/2 =sin(2wx-π/6)+1/2 (1) 最小正周期T=2π/2w=π w=1 (2) x∈[0,2π/3] 2x-π/6∈[-π/6, 7π/6]sin(2x-π/6)∈[-1/2...

解：（Ⅰ）∵函数f(x)=sin2ωx+√3sinωxsin(ωx+π2)=sin2ωx+√3sinωx • cosωx =1-cos2ωx2+√3sin2ωx2=12+sin（2ωx-π6），且它的周期等于π，∴2π2ω=π，∴ω=1，∴f（x）=12+sin（2x-π6）．（Ⅱ）由 2kπ+π2≤2x-π6≤2kπ+3π2，k∈z，...

x）=sin2ωx+3sinωxcosωx，则f(x)＝1?cos2ωx2+32sin2ωx=sin(2ωx?π6)+12，∵T＝2π2ω＝π，∴ω=1．（Ⅱ）f(x)＝sin(2x?π6)+12，∵0≤x≤π2，∴?π6≤2x?π6≤5π6．∴0≤in(2x?π6)+12≤32．∴函数f（x）在区间[0，π2]上的取值范围[0，32]．

f(x)=sin²ωx+根号3sinωxsin[ωx+π/2]=1/2-(1/2)cos2wx+√3sinwxcoswx =(√3/2)sin2wx-(1/2)cos2wx+1/2 =sin(2wx-π/6)+1/2 (1) 最小正周期T=2π/2w=π w=1 (2)x∈[0,2π/3] 2x-π/6∈[-π/6, 7π/6]sin(2x-π/6)∈[-1/2, 1]...

f(x)=sin(2x-π/6)+1/2 0<=x<=2π/3 -π/6<=2x-π/6<=7π/6 sin在(2kπ-π/2,2kπ+π/2)是增函数 在(2kπ+π/2,2kπ-π/2)是减函数 所以-π/6<=2x-π/6<=π/2 -1/2<=sin(2x-π/6)<=1 π/2<=2x-π/6<=7π/6 -1/2<=sin(2x-π/6)<=1 所以-1...