已知f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx.(1)求函数f(x)的最小正周期;(2)当x∈[π/2,π],函数f(x)=0的根.
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/06 07:20:49
![已知f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx.(1)求函数f(x)的最小正周期;(2)当x∈[π/2,π],函数f(x)=0的根.](/uploads/image/z/2485218-66-8.jpg?t=%E5%B7%B2%E7%9F%A5f%28x%29%3Dcos%283x%2F2%29cos%28x%2F2%29-sin%283x%2F2%29sin%28x%2F2%29-2sinxcosx.%EF%BC%881%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%EF%BC%9B%EF%BC%882%EF%BC%89%E5%BD%93x%E2%88%88%5B%CF%80%2F2%2C%CF%80%5D%2C%E5%87%BD%E6%95%B0f%28x%29%3D0%E7%9A%84%E6%A0%B9.)
已知f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx.(1)求函数f(x)的最小正周期;(2)当x∈[π/2,π],函数f(x)=0的根.
已知f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx.
(1)求函数f(x)的最小正周期;
(2)当x∈[π/2,π],函数f(x)=0的根.
已知f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx.(1)求函数f(x)的最小正周期;(2)当x∈[π/2,π],函数f(x)=0的根.
f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx=
=[cos(3x/2+x/2)]-sin2x
=cos2x-sin2x
=√2[√2/2cos2x-√2/2sin2x]
=√2cos(2x+π/4)
(1)
最小正周期为T=2π/2=π
(2)π/2≤x≤π
5π/4≤2x+π/4≤2π+π/4
f(x)=0可化为;
cos(2x+π/4)=0
2x+π/4=3π/2;
x=5π/8
f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx=cos(3x/2+x/2)-sin2x
=cos2x-sin2x=根号2cos(2x+π/4)
所以f(x)的最小正周期=2π/2=π;
x∈[π/2,π],2x∈[π,2π],2x+π/4∈[5π/4,9π/4],
f(x)=0; cos(2x+π/4)=0;
2x+π/4=3π/2
x=5π/8
f(x)=cos(3x/2+x/2)-sin2x
=cos2x-sin2x
=√2(√2/2cos2x-√2/2sin2x)
=√2(cos(π/4+2x))
(1) T=2π/2=π
(2) x∈[π/2,π] π=<2x<2π 5π/4=<π/4+2x<=9π/4
f(x)=√2cos(π/4+2x)=0
π/4+2x=3π/2
x=5π/8
f(x)=cos(3x/2+x/2)=cos2x
∴最小正周期是2π/2=π;在[π/2,π]上,若f(x)=0,则x=π/4
f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx.
=cos(3x/2+x/2)-sin2x
=cos2x-sin2x
=√2cos(2x+π/4)
(1)函数f(x)的最小正周期是2π/2=π
(2) f(x)=0 2x+π/4=kπ+π/2 x=(kπ+π/4)...
全部展开
f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx.
=cos(3x/2+x/2)-sin2x
=cos2x-sin2x
=√2cos(2x+π/4)
(1)函数f(x)的最小正周期是2π/2=π
(2) f(x)=0 2x+π/4=kπ+π/2 x=(kπ+π/4)/2
π/2<=(kπ+π/4)/2<=π
π<=kπ+π/4<=2π 3/4<=k<=7/4 k=1 x=5π/8
当x∈[π/2,π],函数f(x)=0的根是:x=5π/8
收起