已知tana/2=根号5求(1+sinα-cosα)/(1+sinα+cosα)
2个回答

因为1=[sin(α/2)]^2+[cos(α/2)]^2, sinα=2sin(α/2)cos(α/2),

cosα=[cos(α/2)]^2-[sin(α/2)]^2

(1+sinα-cosα)/(1+sinα+cosα)

={[sin(α/2)+cos(α/2)]^2-[[cos(α/2)]^2-[sin(α/2)]^2]} / {[sin(α/2)+cos(α/2)]^2+[[cos(α/2)]^2-[sin(α/2)]^2]}

={[sin(α/2)]^2+sin(α/2)cos(α/2)} / {[cos(α/2)]^2+sin(α/2)cos(α/2)}

上下同除以[cos(α/2)]^2, 得到

={[tan(α/2)]^2+tan(α/2)} / [tan(α/2)+1]

=(5+√5)/(1+√5)

=√5

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