f(x)=2sinxcos(3π/2+x)+√3sin(π+x)*cosx+sin(π/2-x)*cosx-1/2(1)求最小正周期和值域(2)写出函数f(x)取得最大值时x的集合(3)求f(x)的单调递增区间
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![f(x)=2sinxcos(3π/2+x)+√3sin(π+x)*cosx+sin(π/2-x)*cosx-1/2(1)求最小正周期和值域(2)写出函数f(x)取得最大值时x的集合(3)求f(x)的单调递增区间](/uploads/image/z/10892217-57-7.jpg?t=f%28x%29%3D2sinxcos%283%CF%80%2F2%2Bx%29%2B%E2%88%9A3sin%28%CF%80%2Bx%29%2Acosx%2Bsin%28%CF%80%2F2-x%29%2Acosx-1%2F2%281%29%E6%B1%82%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E5%92%8C%E5%80%BC%E5%9F%9F%282%29%E5%86%99%E5%87%BA%E5%87%BD%E6%95%B0f%28x%29%E5%8F%96%E5%BE%97%E6%9C%80%E5%A4%A7%E5%80%BC%E6%97%B6x%E7%9A%84%E9%9B%86%E5%90%88%283%29%E6%B1%82f%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4)
f(x)=2sinxcos(3π/2+x)+√3sin(π+x)*cosx+sin(π/2-x)*cosx-1/2(1)求最小正周期和值域(2)写出函数f(x)取得最大值时x的集合(3)求f(x)的单调递增区间
f(x)=2sinxcos(3π/2+x)+√3sin(π+x)*cosx+sin(π/2-x)*cosx-1/2
(1)求最小正周期和值域
(2)写出函数f(x)取得最大值时x的集合
(3)求f(x)的单调递增区间
f(x)=2sinxcos(3π/2+x)+√3sin(π+x)*cosx+sin(π/2-x)*cosx-1/2(1)求最小正周期和值域(2)写出函数f(x)取得最大值时x的集合(3)求f(x)的单调递增区间
(1)
f(x)=2sinxsinx+√3(-sinx)cosx+cosxcosx-1/2
=(1-cos2x)-√3/2sin2x+1/2(1+cos2x)-1/2
=1-1/2cos2x-√3/2sin2x+1/2-1/2
=1-sin(2x+π/6)
T=2π/2=π
y(MAX)=2
y(min)=0
值域为;
[0,2]
(2)
当sin(2x+π/6)=-1
2x+π/6= - π/2+2kπ==>x= - π/3+kπ时,函数取最大值,
最大值的集合:
{x| x= - π/3+kπ,k∈z}
当
当sin(2x+π/6)=1
2x+π/6= π/2+2kπ==>x= π/6+kπ时,函数取最小值,
最小值的集合:
{x| x=π/6+kπ,k∈z}
(3)
把中间变量 :2x+π/6代入到标准函数的单调减区间中去解出单调增区间的过程:
π/2+2kπ≤2x+π/6≤3π/2+2kπ
π/6+kπ≤ x ≤2π/3+kπ
所以原函数的间调增区间是:
[π/6+kπ ,2π/3+kπ]