Question

已知a=(sin θ,1),b=(1,cos θ),c=(0,3),-π/2< θ<π/2 求|a+b|的取值范围

Answer 1

|a+b|^2=sin^2θ+1+cos^2θ+1+2ab
=3+2(sinθ+cosθ)
=3+2根号2sin(θ+π/4)
因为-π/2< θ<π/2
所以-π/4<θ+π/4<3π/4
所以-根号2/2<sin(θ+π/4)≤1
所以|a+b|^2的范围是（1，3+2根号2】
所以|a+b|的范围是（1，根号2+1】

Answer 2

|a+b|=√[(sin θ-1)²+(1-cos θ)²]=√(3-2sin θ-2cos θ)=√[3-2√2sin (θ+π/4)]
-π/2< θ<π/2
当θ=π/4时,|a+b|有最小值=√[3-2√2sin (π/4+π/4)]=√(3-2√2)=√2-1
当θ=-π/2时,|a+b|=√[3-2√2sin (-π/2+π/4)]=√(3+2)=√5
|a+b|的取值范围是[√2-1,√5]