Question

已知cosα=3/5，sin(α-β)=-√5/5，且α∈（0，π/2），α-β∈（-π/2，0）求cosβ

Answer 1

α∈（0，π/2），sinα=4/5,
α-β∈（-π/2，0）,
sin(α-β)=-√5/5,
cos(α-β)=2√5/5,
cosβ=cos(-β)=cos(α-β-α)=cos(α-β)cosα+sin(α-β)sinα
=2√5/5*3/5+(-√5/5)*4/5
=2√5/25.

Answer 2

(sinα)^2+(cosα)^2=1
(sinα)^2=1-(cosα)^2=1-(3/5)^2=1-9/25=16/25
α∈(0，π/2)，sinα=4/5
[cos(α-β)]^2=1-[sin(α-β)]=1-1/5=4/5
α-β∈(-π/2，0)，cos(α-β)=2√5/5
cosβ=cos(-β)=cos(α-β-α)=cos(α-β)cosα+sin(α-β)sinα=(2√5/5)x(3/5)+(-√5/5)x(4/5)=2√5/25

Answer 3

cosβ=cos(-β)=cos(α-β-α)=cos(α-β)cosα+sin(α-β)sinα
sin(α-β)=-√5/5,α-β∈（-π/2，0）cos(α-β)=2√5/5
cosα=3/5，α∈（0，π/2），sinα=4/5,
cosβ=2√5/5 * 3/5+(-√5/5 * 4/5)=2√5/25

Answer 4

a=2b
cosB=0.4*√5