Question

已知a∈(0,π/2),且sin^2a-sinacosa-2cos^2=0,求sin(a+π/4)/(sin2a+cos2a+1)=?

Answer 1

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

Answer 2

已知a∈(0,π/2),且sin^2a-sinacosa-2cos^2=0,求sin(a+π/4)/(sin2a+cos2a+1)=?

sin^2a-sinacosa-2cos^2=0,
(sina-2cosa)(sina+cosa)=0

a∈(0,π/2),
sina+cosa≠0
sina-2cosa=0
sina=2cosa
(sina)^2=(2cosa)^2
1-(cosa)^2=4(cosa)^2
5(cosa)^2=1
cosa=1/√5

sin(a+π/4)/(sin2a+cos2a+1)
=(sina*cosπ/4+cosa*sinπ/4)/[2sinacosa+2(cosa)^2-1+1]
=(√2/2)(sina+cosa)/2cosa(sina+cosa)
=√2/4cosa
=√2/4(1/√5)
=√10/4