0＜a,β＜π/2,且cosa=根号2/10,cosβ=2根号5/51.求sin(a+β)2.求tan(a+β)

Question

0＜a,β＜π/2,且cosa=根号2/10,cosβ=2根号5/5 1.求sin(a+β) 2.
求tan(a+β)

Answer 1

因为sin^2a+cos^a=1所以sina=±根号（1-cosa）又因为0＜a,β＜π/2所以sinα>0,sinβ>0
所以sinα=7根号2/10,sinβ=根号5/5
sin(α+β)=sinα·cosβ+cosα·sinβ=3根号10/10
cos(α+β）=根号10/10
tan(α+β）=sin(α+β）/cos(α+β）=3
打字不容易,