----(1+sinx-cosx)/(1+sinx+cosx) = [(1-cosx)+sinx]/[(1+cosx)+sinx] = {2*[sin(x/2)]^2+2sin(x/2)cos(x/2)}/{2*[cos(x/2)]^2+2sin(x/2)cos(x/2)} = sin(x/2)/cos(x/2) = tan(x/2)(1-cosx)=2*[sin(x/2)]^2是怎么来的？？？2*[sin(x/2)]^2+2sin(x/2)cos(x/2)]=sin(x/2)是怎么来的饿。。。。不明白求解释、cos2x=(cosx)^2-(sinx)^2=1-2(sinx)^2;(1+sinx-cosx)/(1+sinx+cosx) = [(1-cosx)+sinx]/[(1+cosx)+sinx] = {2*[sin(x/2)]^2+2sin(x/2)cos(x/2)}/{2*[cos(x/2)]^2+2sin(x/2)cos(x/2)} =[(2sinx/2+2cosx/2) * sinx/2] / [(2sinx/2+2cosx/2) * cosx/2]= sin(x/2)/cos(x/2) = tan(x/2)明白了。。。谢谢