已知cos(π/4+x)=3/5,5/4π<x<7π/4,求(sin2x+2sin²x)/(1-tanx)

4个回答

  • cos(π/4+x)=3/5,3/2π<π/4+x<2π,

    所以sin(π/4+x)=-4/5,

    Sin2x=-cos(π/2+2x)

    =-cos[2(π/4+x)]

    =-[1-2sin²(π/4+x)]=7/25.

    sinx+cosx=√2* sin(π/4+x)=-4√2/5,

    cosx –sinx=√2* cos(π/4+x)=3√2/5,

    (sin2x+2(sinx)^2)/(1-tanx)

    =[2sinx(sinx+cosx)]/(1-sinx/cosx)……分子分母同乘以cosx可得下式

    =[2sinx cosx (sinx+cosx)]/( cosx -sinx)

    =[sin2x (sinx+cosx)]/( cosx -sinx)

    =[7/25*(-4√2/5)]/(3√2/5)

    =-28/75.