化简 (21 15:57:26)（sin4α）/ [4sin2（π/4 +α）tan（π/4 -α）]=

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化简 (21 15:57:26)（sin4α）/ [4sin2（π/4 +α）tan（π/4 -α）]=

化简 (21 15:57:26)（sin4α）/ [4sin2（π/4 +α）tan（π/4 -α）]=
化简 (21 15:57:26)
（sin4α）/ [4sin2（π/4 +α）tan（π/4 -α）]=

化简 (21 15:57:26)（sin4α）/ [4sin2（π/4 +α）tan（π/4 -α）]=
sin(π/4+α)=cos[π/2-(π/4+α)]=cos(π/4-α)
tan(π/4-α)=sin(π/4-α)/cos(π/4-α)
所以分母=4cos²(π/4-α)*sin(π/4-α)/cos(π/4-α)
=4sin(π/4-α)cos(π/4-α)
=2sin[2(π/4-α)]
=2sin(π/2-2α)
=2cos2α
分子=sin4α=2sin2αcos2α
所以原式=sin2α

tan（π/4 -α）=cot[π/2-(π/4 -α)]=cot(π/4 +α)
所以原式=（sin4α）/ [4sin2（π/4 +α）cot(π/4 +α)]
=（sin4α）/ [4sin2（π/4 +α） *cos(π/4 +α)/sin(π/4 +α)]
=（sin4α）/ [4sin（π/4 +α） *cos(π/4 +α)]<...

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tan（π/4 -α）=cot[π/2-(π/4 -α)]=cot(π/4 +α)
所以原式=（sin4α）/ [4sin2（π/4 +α）cot(π/4 +α)]
=（sin4α）/ [4sin2（π/4 +α） *cos(π/4 +α)/sin(π/4 +α)]
=（sin4α）/ [4sin（π/4 +α） *cos(π/4 +α)]
=（sin4α）/ [2sin(π/2 +2α)]
=（sin4α）/ (2cos2α)
=2sin2αcos2α/(2cos2α)
=sin2α
分数给我吧？O(∩_∩)O谢谢！

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