a=(2cosβ,2sinβ）,β属于（π/2,π）b=(0,-1)|a|= 2|b| = 1设a,b夹角为 xa.b = (2cosβ,2sinβ).(0,-1)|a||b|cosx = -2sinβcosx = -sinβcosx = -cos(π/2- β)= cos[π+(π/2- β)]= cos(3π/2- β) x= 3π/2- ...