已知向量a=(sinΘ,2),b=(cosΘ,1)且a‖b,其中Θ∈(0,π/2)

1
a∥b,即：sinθ/cosθ=2
即：sinθ=2cosθ,即：4cosθ^2+cosθ^2=1
即：cosθ^2=1/5,θ∈(0,π/2)
故：cosθ=√5/5,sinθ=2√5/5
2
β∈(0,π/2),即：-β∈(-π/2,0)
故：θ-β∈(-π/2,π/2)
故：cos(θ-β)=4/5
cosβ=cos(θ-(θ-β))
=cosθcos(θ-β)+sinθsin(θ-β)
=(√5/5)(4/5)+(2√5/5)(3/5)
=2√5/5