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化简[sin(π/2+α)*cos(π/2-α)]/cos(π+α)+[sin(π-α)*cos(π/2+α)]/sin(π+α)
人气:473 ℃ 时间:2020-04-08 13:28:20
解答
[sin(π/2+α)*cos(π/2-α)]/cos(π+α)+[sin(π-α)*cos(π/2+α)]/sin(π+α)
=cosαsinα/(-cosα)+sinα(-sinα)/(-sinα)
=-sinα+sinα
=0
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