已知函数f(x)=4sinx/2cos(x/2-π /6)求单调递减区间及对称中心以及在区间[π/2,4π/3]上的值域

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  • sinα*sinβ=[cos(α-β)-cos(α+β)]/2

    cos(x/2-π/6)=sin(x/2-π/6+π/2)=sin(x/2+π/3)

    f(x)=4sinx/2sin(x/2+π/3)

    =4*[cosπ/3-cos(x+π/3)]/2

    =2*[1/2-cos(x+π/3)]

    =-2cos(x+π/3)+1

    =2cos(x-2π/3)+1

    x-2π/3∈[2kπ,π+2kπ]单调减,即减区间为[2π/3+2kπ,5π/3+2kπ]

    (x-2π/3∈[-π+2kπ,2kπ]单调增,即增区间为[-π/3+2kπ,2π/3+2kπ])

    对称中心x-2π/3=π/2+kπ 即x=π/6+kπ k∈z

    区间[π/2,4π/3],x-2π/3∈[-π/6,2π/3] cos(x-2π/3)∈[-1/2,1] 2cos(x-2π/3)∈[-1,2]

    在区间[π/2,4π/3],值域为[0,3]