已知函数f(x)=2cosxsin(x+π⼀3)-根号3sin^2x+sinxcosx 当x∈[0,π⼀4]时,f(x)的值域

2024-12-13 17:40:35
推荐回答(1个)
回答1:

f(x)=2cosxsin(x+π/3)-√3*(sinx)^2+sinxcosx
=2cosx[sinxcos(π/3)+cosxsin(π/3)]-√3*(sinx)^2+sinxcosx
=sinxcosx+√3*(cosx)^2-√3*(sinx)^2+sinxcosx
=2sinxcosx+√3*[(cosx)^2-(sinx)^2]
=sin2x+√3*cos2x
=2sin(2x+π/3)

x∈[0,π/4]
2x∈[0,π/2]
2x+π/3∈[π/3,5π/6]
sin(2x+π/3)∈[1/2,1]
2sin(2x+π/3)∈[1,2]

故f(x)的值域是[1,2]

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