Question

已知sinα=4/5,α∈(π/2,π),cosβ=-5/13,β∈(π，3π/2)求

已知sinα=4/5,α∈(π/2,π),cosβ=-5/13,β∈(π，3π/2)求sin(α＋β),cos(α－β)

Answer 1

sina=4/5 兀/2<a<兀 确定符号 cosa=-3/5
cosb=-5/13 兀<b<3兀/2 确定符号 sinb=-12/13
sin(a+b)
=sinacosb+cosasinb=(4/5)*(-5/13)-(3/5)*(-12/13)=16/65

cos(a-b)
=cosacosb+sinasinb==(-3/5)*(-5/13)+(4/5)*(-5/13)=-1/13 希望看的明白

Answer 2

∵α∈(π/2,π),β∈(π，3π/2)
∴cosα=-√[1-(sinα)^2]=-√[1-(4/5)^2]=-3/5
sinβ=-√[1-(cosβ)^2]=-√[1-(-5/13)^2]=-12/13
sin(α＋β)=sinαcosβ+cosαsinβ=4/5*(-5/13)+(-3/5)*(-12/13)=16/65
cos(α-β)=cosαcosβ+sinαsinβ=(-3/5)*(-5/13)+4/5*(-12/13)=-33/65

Answer 3

sina=4/5兀/2<a<兀cosa=-3/5
cosb=-5/13 兀<b<3兀/2
sinb=-12/13
sin(a+b)
=sinacosb+cosasinb=(4/5)*(-5/13)-(3/5)*(-12/13)=16/65
cos(α－β)=cosacosb+sinasinb=33/65
望采纳