已知a∈(0,π/2),且sin^2a-sinacosa-2cos^2=0,求sin(a+π/4)/(sin2a+cos2a+1)=?
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sin^2a-sinacosa-2cos^2a=0
(sina+cosa)(sina-2cosa)=0
sina+cosa=0,或sina-2cosa＝0
sina+cosa=0（舍去）,
sina=2cosa
cos^2a+sin^2a=1
sina=√5/5或sina=－√5/5（舍去）
sina=√5/5
sin(a+π/4)/(sin2a+cos2a+1)
=sin(a+π/4)/(sin2a+cos2a+1)
=sin(a+π/4)/[(sina+cosa)^2+cos2a]
=sin(a+π/4)/[(sina+cosa)^2+(cos^2a-sin^2a)]
=sin(a+π/4)/(sina+cosa)2cos2a
=sin(a+π/4)/(sina+cosa)sina
=√2/2(sina+cosa)/sina+cosa)sina
=√2/2sina
=√2/2*√5/5＝√10/10
(sina+cosa)(sina-2cosa)=0
sina+cosa=0,或sina-2cosa＝0
sina+cosa=0（舍去）,
sina=2cosa
cos^2a+sin^2a=1
sina=√5/5或sina=－√5/5（舍去）
sina=√5/5
sin(a+π/4)/(sin2a+cos2a+1)
=sin(a+π/4)/(sin2a+cos2a+1)
=sin(a+π/4)/[(sina+cosa)^2+cos2a]
=sin(a+π/4)/[(sina+cosa)^2+(cos^2a-sin^2a)]
=sin(a+π/4)/(sina+cosa)2cos2a
=sin(a+π/4)/(sina+cosa)sina
=√2/2(sina+cosa)/sina+cosa)sina
=√2/2sina
=√2/2*√5/5＝√10/10
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