作业帮已知(π/4+α)=-1/2,试求式子(sin2α-2cos^2α)/(1+tanα)的值.已知tan(π/4+α)=-1/2,试求式子(sin2α-2cos^2α)/(1+tanα)的值
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已知(π/4+α)=-1/2,试求式子(sin2α-2cos^2α)/(1+tanα)的值.已知tan(π/4+α)=-1/2,试求式子(sin2α-2cos^2α)/(1+tanα)的值
已知(π/4+α)=-1/2,试求式子(sin2α-2cos^2α)/(1+tanα)的值.
已知tan(π/4+α)=-1/2,试求式子(sin2α-2cos^2α)/(1+tanα)的值
已知(π/4+α)=-1/2,试求式子(sin2α-2cos^2α)/(1+tanα)的值.已知tan(π/4+α)=-1/2,试求式子(sin2α-2cos^2α)/(1+tanα)的值
(sin2α-2cos^2α)/(1+tanα)＝(2sinαcosα-2cosα^2)/(1+tanα)
=(2sinαcosα-2cosα^2)/[(1+tanα)(sinα^2+cosα^2)]
=(2tanα-2)/[(1+tanα)(tanα^2+1)]
因为tan(π/4+α)=-1/2＝(1+tanα)/(1-tanα),解出tanα,代入上式即可.
tan(π/4 +α) =(tanπ/4+tanα)/(1-tanπ/4*tanα) =(1+tanα)/(1-tanα)=-1/2 所以tanα=-3 sinα/cosα=tanα=-3 sinα=-3cosα 代入(sinα)^2+(cosα)^2=1 (sinα)^2=9/10 (cosα)^2=1/10 sinα/cosα=-3<0 所以sinαcosα<0 ...
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tan(π/4 +α) =(tanπ/4+tanα)/(1-tanπ/4*tanα) =(1+tanα)/(1-tanα)=-1/2 所以tanα=-3 sinα/cosα=tanα=-3 sinα=-3cosα 代入(sinα)^2+(cosα)^2=1 (sinα)^2=9/10 (cosα)^2=1/10 sinα/cosα=-3<0 所以sinαcosα<0 所以sinαcosα=-根号(sinα)^2(cosα)^2=-3/10 sin2α=2sinαcosα=-3/5 所以原式=(-3/5-2*1/10)/(1-3)=2/5
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