已知a=(sin θ,1),b=(1,cos θ),c=(0,3),-π/2扫码下载作业帮搜索答疑一搜即得
问题描述：
已知a=(sin θ,1),b=(1,cos θ),c=(0,3),-π/2
扫码下载作业帮
搜索答疑一搜即得
答
|a+b|=√[(sin θ-1)²+(1-cos θ)²]=√(3-2sin θ-2cos θ)=√[3-2√2sin (θ+π/4)]
-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]
答
|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所以-π/4所以-根号2/2
所以|a+b|的范围是（1,根号2+1】