Question

已知向量a（cosθ，sinθ）向量b=（根号3，1）当向量a⊥向量b，求tan2θ（2）/向量a+向量b/（模）的最大值

Answer 1

(1）当向量a⊥向量b，=>
√3cosθ+sinθ=2sin(θ+π/6)=0
=>θ+π/6=2kπ
=>θ=2kπ-π/6
=>tan2θ=tan(4kπ-π/3)=-√3
(2)la+bl²
=a²+2a.b+b²
=1+2+2sin(θ+π/6)
=3+2sin(θ+π/6)
=>1≤la+bl²≤5
=>1≤la+bl≤√5

追问

这一步a²+2a.b+b²是怎么变成1+2+2sin(θ+π/6)的？为什么2ab变成了2sin(θ+π/6)？

追答

a²=cos²θ+sin²θ=1 同理：b²=2
ab=√3cosθ+sinθ=2sin(θ+π/6)
(2)la+bl²
=a²+2a.b+b²
=1+2+4sin(θ+π/6)
=3+4sin(θ+π/6)
=>0≤la+bl²≤7
=>-√7≤la+bl≤√7