(1+sinx-cosx)/(1+sinx+cosx)化简过程,
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(1+sinx-cosx)/(1+sinx+cosx)化简过程,
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(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)
再问: (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、
再答: 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)
再问: 明白了。。。谢谢
(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)
再问: (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、
再答: 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)
再问: 明白了。。。谢谢
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