求证（2cosθ-sin2θ)/(2cosθ+sin2θ)=tan(π/4-θ/2)右边应该是tan(π/4-θ/2)^2
问题描述：
求证（2cosθ-sin2θ)/(2cosθ+sin2θ)=tan(π/4-θ/2)
右边应该是tan(π/4-θ/2)^2
答
（2cosθ-sin2θ)/(2cosθ+sin2θ)分子分母同时提取2cosθ，等式左边等于：
（1-sinθ）/(1+sinθ)=((cosθ/2)^2+(sinθ/2)^2-2sinθ/2*cosθ/2)/((cosθ/2)^2+(sinθ/2)^2+2sinθ/2*cosθ/2)=((cosθ/2+sinθ/2)/(cosθ/2-sinθ/2))^2=((1+tanθ/2)/(1-tanθ/2))^2=tan(π/4-θ/2)^2；得证
答
（2cosθ-sin2θ)/(2cosθ+sin2θ)=2cosθ(1-sinθ)/[2cosθ(1+sinθ)]=(1-sinθ)/(1+sinθ)tan(π/4-θ/2)=sin(π/4-θ/2)/cos(π/4-θ/2)=(cosθ/2-sinθ/2)/(sinθ/2+cosθ/2)=根号(cosθ/2-sinθ/2)^2/(sinθ/2+...