于知函数f(X)=sin²ωx+根号3sinωxsin(ωx+π/2(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]上的取值范围
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![于知函数f(X)=sin²ωx+根号3sinωxsin(ωx+π/2(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]上的取值范围](/uploads/image/z/4333324-4-4.jpg?t=%E4%BA%8E%E7%9F%A5%E5%87%BD%E6%95%B0f%28X%29%3Dsin%26sup2%3B%CF%89x%2B%E6%A0%B9%E5%8F%B73sin%CF%89xsin%28%CF%89x%2B%CF%80%2F2%28%CF%89%3E0%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E4%B8%BA%CF%80.%EF%BC%881%EF%BC%89%E6%B1%82%CF%89%E7%9A%84%E5%80%BC%EF%BC%9B%EF%BC%882%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%5B0%2C2%CF%80%2F3%5D%E4%B8%8A%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
于知函数f(X)=sin²ωx+根号3sinωxsin(ωx+π/2(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]上的取值范围
于知函数f(X)=sin²ωx+根号3sinωxsin(ωx+π/2(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]上的取值范围
于知函数f(X)=sin²ωx+根号3sinωxsin(ωx+π/2(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]上的取值范围
已知函数f(x)=根号下3 sinωxcosωx-cos 2;ωx(ω0)的周期为π/=2cosωx[sinωxcos(π/6)-cosωxsin(π/6)] 即,f(x)=2cosωx*
sin²ωx=1-cos2ωx平方
sin(ωx+π/2)=cosωx
f(x)=3/2sin2ωx-cos2ωx+1
f(x)=开方【(3/2)平方+1的平方】sin(2ωx+a)+1(a为常数
所以2ω=2得ω=1
取值范围自已求,打字不便
f(x)=sin²ωx+√3sinωxcosωx
=1/2-1/2cos2ωx+√3/2sin2ωx
=1/2+sin2ωxcosπ/6-cos2ωxsinπ/6
=sin(2ωx+π/6)+1/2
根据题意
2π/2ω=π
ω=1
f(x)=sin(2x+π/6)+1/2
0≤x≤2π/3
0≤2x≤4π/3
π/6≤2x+π/6≤3π/2
所以sin(2x+π/6)∈[-1,1]
所以f(x)∈[-1/2,3/2]
f(x)=sin^2ωx+√3sinωxsin(ωx+π/2)
=sin^2ωx+√3sinωxcosωx
=1-cos^2wx+√3/2*sin2wx
=1-(cos2wx+1)/2+√3/2*sin2wx
=1/2+(√3/2*sin2wx-1/2*cos2wx)
=1/2+sin(2wx-π/6)
1)
最小正周期=2π/2w=π
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f(x)=sin^2ωx+√3sinωxsin(ωx+π/2)
=sin^2ωx+√3sinωxcosωx
=1-cos^2wx+√3/2*sin2wx
=1-(cos2wx+1)/2+√3/2*sin2wx
=1/2+(√3/2*sin2wx-1/2*cos2wx)
=1/2+sin(2wx-π/6)
1)
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]
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