已知tan(π/4+α)=1/5,求(sin2α-sin²α)/(1-cos2α)的值
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已知tan(π/4+α)=1/5,求(sin2α-sin²α)/(1-cos2α)的值
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由已知可得 tanα=tan[(α+π/4)-π/4]=[tan(α+π/4)-tan(π/4)] / [1+tan(α+π/4)tan(π/4)]
=(1/5-1)/(1+1/5)= -2/3 ,
所以 [sin(2α)-(sinα)^2] / [1-cos(2α)]
=sinα(2cosα-sinα) / [2(sinα)^2]
=(2cosα-sinα) / (2sinα)
=1/tanα-1/2
= -3/2-1/2
= -2 .
=(1/5-1)/(1+1/5)= -2/3 ,
所以 [sin(2α)-(sinα)^2] / [1-cos(2α)]
=sinα(2cosα-sinα) / [2(sinα)^2]
=(2cosα-sinα) / (2sinα)
=1/tanα-1/2
= -3/2-1/2
= -2 .
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