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sin(540-α)/tan(α-180)*cot(-α-360)/tan(900-α)*cos(720-α)/sin(-α-360)
人气:197 ℃ 时间:2019-10-09 12:09:42
解答
原式=sin(180º-α)/tan(180º+α)]·cot(-α)]/tan(180º-α)·cos(-α)/sin(-α)
=sinα/tanα·(-cotα)/(-tanα)·cosα/(-sinα)
=sinα·(cosα/sinα)·cot²α·(-cotα)
=-cosα·cot³α
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