作业帮已知函数f(x)= √3sinωxcosωx-cos^2ωx+1/2(ω>0,x∈R)的最小正周期为π/21、求f(2π/3)的值,并求出函数f(x)的图像的对称中心的坐标2、当x∈[π/3,π/2]时,求函数f(x)的单调递增区间注：π=pi
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/29 08:51:43
![已知函数f(x)= √3sinωxcosωx-cos^2ωx+1/2(ω>0,x∈R)的最小正周期为π/21、求f(2π/3)的值,并求出函数f(x)的图像的对称中心的坐标2、当x∈[π/3,π/2]时,求函数f(x)的单调递增区间注：π=pi](/uploads/image/z/3857411-11-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D+%E2%88%9A3sin%CF%89xcos%CF%89x-cos%5E2%CF%89x%2B1%2F2%28%CF%89%3E0%2Cx%E2%88%88R%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E4%B8%BA%CF%80%2F21%E3%80%81%E6%B1%82f%282%CF%80%2F3%29%E7%9A%84%E5%80%BC%2C%E5%B9%B6%E6%B1%82%E5%87%BA%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%9B%BE%E5%83%8F%E7%9A%84%E5%AF%B9%E7%A7%B0%E4%B8%AD%E5%BF%83%E7%9A%84%E5%9D%90%E6%A0%872%E3%80%81%E5%BD%93x%E2%88%88%5B%CF%80%2F3%2C%CF%80%2F2%5D%E6%97%B6%2C%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%E6%B3%A8%EF%BC%9A%CF%80%3Dpi)
已知函数f(x)= √3sinωxcosωx-cos^2ωx+1/2(ω>0,x∈R)的最小正周期为π/21、求f(2π/3)的值,并求出函数f(x)的图像的对称中心的坐标2、当x∈[π/3,π/2]时,求函数f(x)的单调递增区间注：π=pi
已知函数f(x)= √3sinωxcosωx-cos^2ωx+1/2(ω>0,x∈R)的最小正周期为π/2
1、求f(2π/3)的值,并求出函数f(x)的图像的对称中心的坐标
2、当x∈[π/3,π/2]时,求函数f(x)的单调递增区间
注：π=pi
已知函数f(x)= √3sinωxcosωx-cos^2ωx+1/2(ω>0,x∈R)的最小正周期为π/21、求f(2π/3)的值,并求出函数f(x)的图像的对称中心的坐标2、当x∈[π/3,π/2]时,求函数f(x)的单调递增区间注：π=pi
f(x)= √3sinωxcosωx-cos^2ωx+1/2
=√3/2*(2sinwxcoswx)-cos^2wx+1/2
=√3/2sin2wx-cos^2wx+1/2
=√3/2sin2wx-[(1+cos2wx)/2]+1/2
=√3/2sin2wx-1/2cos2wx
=-cos(π/3+2wx)
因为最小正周期为π/2,所以T=2π/2w=π/2,w=2
f(x)=-cos(π/3+4x)
x=2π/3时,f(x)=1,由正余弦图像知,其中心对称坐标即为其与X轴的交点,所以f(x)的中间对称坐标为：(π/24+kπ/2,0),
(2)f(x)=-cos(π/3+4x),其单调增区间为-π+2kπ≤π/3+4x≤0,即-π/3+1/2kπ ≤ x ≤ -π/12+1/2kπ即[-π/3+1/2kπ,-π/12+1/2kπ]
同理,其单调减区间为：-π/12+1/2kπ ≤ x ≤ π/6+1/2kπ即[-π/12+1/2kπ,π/6+1/2kπ ]
所以x∈[π/3,π/2]时,
x∈[π/3,5π/12]为增函数,在x∈[5π/12,π/2]为减函数.
上面所以的K为整数.