化简sin²(α+π)·cos(π+α)·cot(-α-2π)/tan(π+α)·cos²(-α-π)求答案

显示全部楼层 · 2013-2-18 15:53:15

解sin2(α+π)·cos(π+α)·cot(-α-2π)/tan(π+α)·cos2(-α-π)=sin2(α)·（-cos(α)）·（-cot(α+2π)）/（tan(α)）·cos2(α)=sin2(α)·cos(α)·cot(α+2π)/（tan(α)）·cos2(α)=sin2(α)·cos(α)·cot(α)/（tan(α)）·cos2(α)=sin2(α)·cos(α)/（tan2(α）·cos2(α)=sin2(α)·cos(α)/tan2(α)·cos2(α)=sin&#...

千问 · 2013-2-18 15:53:15

[sin2(α+π)·cos(π+α)·cot(-α-2π)]/[tan(π+α)·cos2(-α-π)]=[sin2α(-cosα)(-cotα)]/[tanαcos2α]=(sin2αcosα*cosα/sinα)/(sinα/cosα*cos2α)
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