已知函数f(x)=2cosxcos(x-π/6)-根号3sin²x+sinxcosx
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已知函数f(x)=2cosxcos(x-π/6)-根号3sin²x+sinxcosx
(1) 求f(x)的最小正周期
(2)当x属于[0,π/2] 时,求f(x)的最小值
(1) 求f(x)的最小正周期
(2)当x属于[0,π/2] 时,求f(x)的最小值
首先化简
f(x)=2cosxcos(x-π/6)-√3sin^2x+sinxcosx
=2cosxcos(x-π/6)-2sinx(√3/2sinx-1/2cosx)
=2cosxcos(x-π/6)-2sinx(sinxcosπ/6-cosxsinπ/6)
=2cosxcos(x-π/6)-2sinxsin(x-π/6)
=2cos(x+x-π/6)
=2cos(2x-π/6)
(1)f(x)的最小正周期T=2π/w=2π/2=π
(2)当x属于[0,π/2] 时,2x-π/6属于[-π/6,5π/6]
所以cos(2x-π/6)属于[-√3/2,1]
所以f(x)的最小值为2*(-√3/2)=-√3
f(x)=2cosxcos(x-π/6)-√3sin^2x+sinxcosx
=2cosxcos(x-π/6)-2sinx(√3/2sinx-1/2cosx)
=2cosxcos(x-π/6)-2sinx(sinxcosπ/6-cosxsinπ/6)
=2cosxcos(x-π/6)-2sinxsin(x-π/6)
=2cos(x+x-π/6)
=2cos(2x-π/6)
(1)f(x)的最小正周期T=2π/w=2π/2=π
(2)当x属于[0,π/2] 时,2x-π/6属于[-π/6,5π/6]
所以cos(2x-π/6)属于[-√3/2,1]
所以f(x)的最小值为2*(-√3/2)=-√3
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