向量与三角函数问题

(1)若a⊥b,则sinθ*1+√3*cosθ=0,2(sinθcosπ/3+cosθsinπ/3)=2sin(θ+π/3)=0,θ+π/3=0,θ=-π/3
(2)a+b=(sinθ+1,√3+cosθ),|a+b|²=5+4sin(θ+π/3),因为θ∈（-π/2,π/2）,所以θ+π/3∈（-π/6,5π/6）,sin(θ+π/3) ∈（-1/2,1],显然当θ=2π/3时,sin(θ+π/3)=1,
|a+b|²=9,|a+b|=3最大.即所求|a+b|最大值为3.