已知tan(α+π/4)=-1/7,α∈(π/2,π),求tanα+(cos2a+1)/[√2cos(α-π/4)-si

1个回答

  • tan(α+π/4)=-1/7,α∈(π/2,π)

    tan(α+π/4)= (tanα + tanπ/4)/(1-tanαtanπ/4) = (tanα + 1)/(1-tanα) = -1/7

    7tanα + 7= - 1+tanα

    tanα = -4/3

    α∈(π/2,π)

    cosα = -1/根号(1+tan^2α)=-1/根号(1+16/9)=-3/5

    sinα=|tanα|/根号(1+tan^2α)=(4/3)/(5/3)=4/5

    cos2α=2cos^2α-1=2*9/25-1= -7/25

    √2cos(α-π/4)=根号2(cosαcosπ/4+sinαsinπ/4)=根号2*根号2/2(-3/4+4/5)=1/5

    sin2α=2sinαcosα=2*4/5*(-3/5)=-24/25

    tanα+(cos2a+1)/[√2cos(α-π/4)-sin2a]

    =[(-4/3)+(-7/25+1)] / [1/5-(-24/25)]

    =(-46/75)/(29/25)

    =-46/87