已知tan(α+π/4)=-1/7,α∈(π/2,π),求tanα+(cos2a+1)/[√2cos(α-π/4)-si
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已知tan(α+π/4)=-1/7,α∈(π/2,π),求tanα+(cos2a+1)/[√2cos(α-π/4)-sin2a]
tan(α+π/4)=-1/7,α∈(π/2,π)
tan(α+π/4)= (tanα + tanπ/4)/(1-tanαtanπ/4) = (tanα + 1)/(1-tanα) = -1/7
7tanα + 7= - 1+tanα
tanα = -4/3
α∈(π/2,π)
cosα = -1/根号(1+tan^2α)=-1/根号(1+16/9)=-3/5
sinα=|tanα|/根号(1+tan^2α)=(4/3)/(5/3)=4/5
cos2α=2cos^2α-1=2*9/25-1= -7/25
√2cos(α-π/4)=根号2(cosαcosπ/4+sinαsinπ/4)=根号2*根号2/2(-3/4+4/5)=1/5
sin2α=2sinαcosα=2*4/5*(-3/5)=-24/25
tanα+(cos2a+1)/[√2cos(α-π/4)-sin2a]
=[(-4/3)+(-7/25+1)] / [1/5-(-24/25)]
=(-46/75)/(29/25)
=-46/87
tan(α+π/4)= (tanα + tanπ/4)/(1-tanαtanπ/4) = (tanα + 1)/(1-tanα) = -1/7
7tanα + 7= - 1+tanα
tanα = -4/3
α∈(π/2,π)
cosα = -1/根号(1+tan^2α)=-1/根号(1+16/9)=-3/5
sinα=|tanα|/根号(1+tan^2α)=(4/3)/(5/3)=4/5
cos2α=2cos^2α-1=2*9/25-1= -7/25
√2cos(α-π/4)=根号2(cosαcosπ/4+sinαsinπ/4)=根号2*根号2/2(-3/4+4/5)=1/5
sin2α=2sinαcosα=2*4/5*(-3/5)=-24/25
tanα+(cos2a+1)/[√2cos(α-π/4)-sin2a]
=[(-4/3)+(-7/25+1)] / [1/5-(-24/25)]
=(-46/75)/(29/25)
=-46/87
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