已知函数y=cos(x- π/6)[√3sin(x-π/6)+cos(x- π/6)],x∈R,求函数y的最大值以及相应

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  • y=cos(x- π/6)[√3sin(x-π/6)+cos(x- π/6)]

    =2cos(x- π/6)[(√3/2)*sin(x-π/6)+(1/2)*cos(x- π/6)] (两角和的正弦公式)

    =2cos(x- π/6)*sin(x-π/6+π/6)

    =2cos(x- π/6)sinx (积化和差公式)

    =sin(x+x-π/6)+sin[x-(x- π/6)]

    =sin(2x-π/6)+1/2

    则当2x-π/6=2kπ+π/2,k∈Z时,函数y有最大值为3/2,

    此时相应的自变量x的集合为:{x | x=kπ+π/3,k∈Z}