化简［sin(2π-α)cos(π+α)cos(π/2+α)cos(11π/2-α)］/［cos(π-α)sin(π+α)sin(-π-α)sin［（...
化简［sin(2π-α)cos(π+α)cos(π/2+α)cos(11π/2-α)］/［cos(π-α)sin(π+α)sin(-π-α)sin［（9π/2）+α］］
人气：268 ℃ 时间：2020-10-01 22:29:19
解答
-sinα*-cosα*-sinα*sinα/-cosα*-sinα*sinα*-cosα
=tanα
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