Question

sinα=2/3， cosβ=-3/4 ，α∈(π/2,π) β∈(π,3π/2)，求cos（α-β）的值

Answer 1

sinα=2/3 α∈(π/2,π) cosα=-根号[1-(2/3)^2]=-根号5/3
cosβ=-3/4 β∈(π,3π/2) sinβ=-根号【1-（-3/4）^2】=-根号7/4
cos（α-β）=cosα*cosβ+sinα*sinβ
=(-根号5/3)*(-3/4)+(2/3)*(-根号7/4)
=根号5/4-根号7/6

Answer 2

因为sinα^2 + cosα^2 = 1 sinβ^2 + cosβ^2 = 1
所以 cosα^2 = 5/9 sinβ^2 = 7/16

因为α∈(π/2,π) β∈(π,3π/2)
所以 cosα = -√5/3 sinβ = -√7/4

cos（α-β）= cosαcosβ + sinαsinβ =(-√5/3)*(-3/4)+ 2/3*(-√7/4) = √5/4 -√7/6