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

题目

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

答案

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.