已知向量a=(cosα,sinα),b=(cosβ,sinβ)（1）当a=5π/6,β=-π/2时,求ab的值

a=(cosα,sinα),b=(cosβ,sinβ)
(1)
ab=cosαcosβ+sinαsinβ
=cos(α-β)
=cos(5π/6-(-π/2))
=-sin5π/6
=-1/2
(2)
ab=cosαcosβ+sinαsinβ
=cos(α-β)=1/3
因为
0＜β＜α＜π/2
所以
0