2道高中的三角函数题1. 已知y=1/2·sin(3x-π/3)-1,x∈[-π/6,π/6]求该三角函数的递增区间、递

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  • sinx在x∈[-π/2+2kπ,π/2+2kπ]单调递增

    则y当3x-π/3∈[-π/2+2kπ,π/2+2kπ]时单调递增

    得x∈[-π/18+2kπ/3,5π/18+2kπ/3]时单调递增

    ∵x∈[-π/6,π/6]

    则[-π/18,π/6]时单调递增

    ∴[-π/6,-π/18]时单调递减(不用求正余弦不是递增就是递减)

    当3x-π/3=π/2 + kπ为对称轴即x=5π/18+kπ/3∵x∈[-π/6,π/6]

    ∴x=-π/18为对称轴

    当3x-π/3= kπ为对称点的x得x=π/9+kπ/3∵x∈[-π/6,π/6]则x=π/9

    所以对称点为(π/9,-1)

    2.y=sin2x+a·cos2x=根号(1+a²)sin(2x+μ)最大幅值为根号(1+a²)

    直线x=-π/8对称,则x=-π/8时y取最值

    ∴y|x=-π/8=-根号2/2+a根号2/2=正负根号(1+a²)

    解得a=-1