已知向量a=(cosα,sinα),b=(cosβ,sinβ)(1)当a=5π/6,β=-π/2时,求ab的值
问题描述:
已知向量a=(cosα,sinα),b=(cosβ,sinβ)(1)当a=5π/6,β=-π/2时,求ab的值
(2)已知ab=1/3,cosα=1/7,0<β<α<π/2,求sinβ的值
答
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