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

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

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

已知tan(π/4+α)=1/2,求(sin2α-cos^2α)/(1+cos2α)的值
tan(π/4+α)=(1+tanα)/(1-tanα)=1/2,得到tanα=-1/3
(sin2α-cos^2α)/(1+cos2α)
=(sin2α-cos^2α)/2cos^2α
=sin2α/2cos^2α -1/2
=tanα-1/2
=-1/3-1/2
=-5/6