已知tan(π/4+α)=1/2,求(sin2α-cos平方α)/(1+cos2α)的值

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已知tan(π/4+α)=1/2,求(sin2α-cos平方α)/(1+cos2α)的值
已知tan(π/4+α)=1/2,求(sin2α-cos平方α)/(1+cos2α)的值

已知tan(π/4+α)=1/2,求(sin2α-cos平方α)/(1+cos2α)的值
tan(α+π/4)=[tanα+tan(π/4)]/[1-tanαtan(π/4)]
=[tanα+1]/[1-tanα]
[tanα+1]/[1-tanα]=1/2
1-tanα=2(tanα+1)
3tanα=-1
tanα=-1/3
(sin2α-cos^2α)/(1+cos2α)
=(2sinαcosα-cos^2α)/(1+2cos^2α-1)
=(2sinαcosα-cos^2α)/(2cos^2α)
=(2sinα-cosα)/(2cosα)
=(2tanα-1)/2
=(-2*1/3-1)/2
=-5/6

tan(π/4+α)=1/2
1+tanα/1-tanα=1/2
tanα=-1/3
(sin2α-cos平方α)/(1+cos2α)
=(2sinαcosα-cos²α)/2cos²α
=tanα-1/2