已知函数f(X)=2sin(X+π/3)-2sinX ,X属于《-π/2,0》
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/21 06:40:14
已知函数f(X)=2sin(X+π/3)-2sinX ,X属于《-π/2,0》
(1)若cosX=√3/3,求函数f(X)的值
(2)求函数f(X)的值域
急.
(1)若cosX=√3/3,求函数f(X)的值
(2)求函数f(X)的值域
急.
f(X)=2sin(X+π/3)-2sinX
=2sinX cos(π/3)+2cosX sin(π/3)-2sinX
=sinX+√3cosX-2sinX
=√3cosX-sinX
=2*[(√3/2)cosX-(1/2)sinX]
=2*[sin(π/3)cosX-cos(π/3)sinX]
=2sin(π/3-X)由此 就解出f(X)
(1)cosX=√3/3,又X属于[-π/2,0],则sinX=-√6/3
由f(X)=√3cosX-sinX=1+√6/3
(2)f(X)=2sin(π/3-X),X属于[-π/2,0],
则π/3-X属于[π/3,5π/6]
画个图便知f(X)的值域为[1,2]
=2sinX cos(π/3)+2cosX sin(π/3)-2sinX
=sinX+√3cosX-2sinX
=√3cosX-sinX
=2*[(√3/2)cosX-(1/2)sinX]
=2*[sin(π/3)cosX-cos(π/3)sinX]
=2sin(π/3-X)由此 就解出f(X)
(1)cosX=√3/3,又X属于[-π/2,0],则sinX=-√6/3
由f(X)=√3cosX-sinX=1+√6/3
(2)f(X)=2sin(π/3-X),X属于[-π/2,0],
则π/3-X属于[π/3,5π/6]
画个图便知f(X)的值域为[1,2]
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