已知向量a=(sinθ,1),b=(1,cosθ),-π/2

已知向量a=(sinθ,1),b=(1,cosθ),-π/2
已知向量a=(sinθ,1),b=(1,cosθ),-π/2<θ<π/2 (1)若a⊥b,求θ （2）求| a+b|的最大值

已知向量a=(sinθ,1),b=(1,cosθ),-π/2
a⊥b =>
a • b = (sinθ) * 1 + 1 * sinθ =0
=>
sinθ = cosθ
又 -π/2
a+b= ( sin(π/4) +1 ,1+cos(π/4))
|a+b|= √{[sin(π/4)+1]^2 + [1+cos(π/4)]^2}
= 1+√2