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化简:cos(π-α)·sin(-π-α) / cos(11/2π-β)·tan(3π+β)
RT、
quickly~
人气:119 ℃ 时间:2020-07-01 18:03:36
解答
cos(π-α)sin(-π-α) / cos(11/2π-β)tan(3π+β)
=cos(π-α)sin[-(π+α)] / cos[4π+(3/2π-β)]tan[2π+(π+β)]
=-cosαsinα / (-sinβ)tanβ
=cosαsinαcosβ / (sinβ)^2
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