Question

已知α∈(0,π/2),β∈(0,π/2),且sinβ=5/13,cos(α+β)=4/5,求sinα

Answer 1

α∈(0,π/2) sinα>0 cosα=12/13 sin 2;α+cos 2;α=1 所以sinα=5/13 β∈(π/2，π) cosβ<0 sinβ=3/5 所以cosβ=-4/5 sin

Answer 2

α∈(0,π/2),β∈(0,π/2), α+β∈(0,π),
sinβ=5/13 cosβ=12/13
cos(α+β)=4/5 sin(α+β)=3/5
sinα=sin[(α+β)-β]
=sin(α+β)cosβ-cos(α+β)sinβ
=3/5*(12/13)-4/5*(5/13)
=16/65

Answer 3

0<β<π/2，cosβ>0
cosβ=√(1-sin²β)=√[1-(5/13)²]==12/13
0<α<π/2，0<α+β<π sin(α+β)>0
sin(α+β)=√[1-cos²(α+β)=√[1-(4/5)²]=3/5
sin(α+β)/cos(α+β)
=(sinαcosβ+cosαsinβ)/(cosαcosβ-sinαsinβ)
=[(12/13)tanα+(5/13)]/[(12/13)-(5/13)tanα]
=(12tanα +5)/(12 -5tanα)=(3/5)/(4/5)=3/4
4(12tanα +5)=3(12 -5tanα)
63tanα=16
tanα=sinα/cosα=16/63
cosα=63sinα/16
sin²α+cos²α=1
sin²α+(63sinα/16)²=1
sin²α+3960sin²α/256=1
sin²α=256/4225
sinα=16/65

Answer 4

sinα=sin(α+β-β)=sin(α+β)cosβ-cos(α+β)sinβ
=(3/5)*(12/13)-(4/5)*(5/13)
=36/65-20/65
=16/65