化简:[sin(2π-α)cos(π+α)cos(π/2 +α)cos(11π/2 -α)]/[cos(π-α)sin(
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化简:[sin(2π-α)cos(π+α)cos(π/2 +α)cos(11π/2 -α)]/[cos(π-α)sin(3π-α)sin(-π-α)sin(9π/2 +α)]
[sin(2π-α)cos(π+α)cos(π/2 +α)cos(11π/2 -α)]/[cos(π-α)sin(3π-α)sin(-π-α)sin(9π/2 +α)]
=[-sinα(-cosα)(-sinα)(-sinα)]/[-cosα sinα sinα cosα]
=-tanα
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=[-sinα(-cosα)(-sinα)(-sinα)]/[-cosα sinα sinα cosα]
=-tanα
如果你认可我的回答,请点击“采纳答案”,祝学习进步!
手机提问的朋友在客户端右上角评价点【评价】,然后就可以选择【满意,问题已经完美解决】了
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