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  • 已知函數(shù)f(x)=cos(x+x/6)-sin(x-2π/3)+sinx+a的最大值為1.求常數(shù)a的值?求使f(x)≥0成立的x的取值范圍

    已知函數(shù)f(x)=cos(x+x/6)-sin(x-2π/3)+sinx+a的最大值為1.求常數(shù)a的值?求使f(x)≥0成立的x的取值范圍
    數(shù)學(xué)人氣:743 ℃時(shí)間:2020-04-18 05:12:44
    優(yōu)質(zhì)解答
    f(x) = cos(x+π/6)-sin(x-2π/3)+sinx+a
    = cos(x+π/6)-{-sin(π+x-2π/3)} + sinx + a
    = cos(x+π/6)+sin(x+π/3)} + sinx + a
    = cos(x+π/6)+cos{π/2-(x+π/3)} + sinx + a
    = cos(x+π/6)+cos{π/6-x} + sinx + a
    = cos(x+π/6)+cos{x-π/6} + sinx + a
    = cosxcosπ/6-sinxsinπ/6 + cosxcosπ/6+sinxsinπ/6 + sinx + a
    = 2cosxcosπ/6 + sinx + a
    = 2(cosxcosπ/6 + sinxsinπ/6) + a
    = 2cos(x-π/6)+a
    -2+a≤2cos(x-π/6)+a≤2+a
    2+a=1
    a=-1
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