您的位置: 首页  >  作业答案  >  其他  >  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］］

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］］

分类: 作业答案 • 2022-01-02 13:20:46
问题描述：

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］］

答

+_+ 不懂

答

只是需要答案哈？不要过程吧

答

-sinα*-cosα*-sinα*sinα/-cosα*-sinα*sinα*-cosα
=tanα

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