已知函数f(x)=cos²x/2-sin²x/2+sin+求f(x)最小正周期

1个回答

  • f(x)=cos²x/2-sin²x/2 + sinx

    =cosx + sinx

    =根号2(sinxcosπ/4+cosxsinπ/4)

    =根号2 sin(x+π/4)

    最小正周期2π

    x0∈(0,π/4)且f(x0)=4根号2分之5,即:

    根号2 sin(x0+π/4)=5/(4根号2)

    sin(x0+π/4)=5/8

    cos(x0+π/4)=根号(1-5^2/8^2))=根号39 /8

    f(x0+π/6)=sin(x0+π/6)=sin[(x0+π/4)-(π/4-π/6)]

    =sin(x0+π/4)cos(π/4-π/6) - cos(x0+π/4)sin(π/4-π/6)

    =5/8(cosπ/4cosπ/6+sinπ/4sinπ/6)-7/8(sinπ/4cosπ/6-cosπ/4sinπ/6)

    =5/8 * 根号2/2 * (根号3/2+1/2) - 根号39 /8* 根号2/2 * (根号3/2-1/2)

    =根号2/32 * [ 5(根号3+1) - 根号39(根号3-1)]

    =根号2/32 * [ 5+5根号3+根号39- 3根号13)]

    = [ 5根号2+5根号6+根号78- 3根号26)]/32