已知向量a＝(Cosα,Sinα),向量b(Cosβ,Sinβ),其中0∠α∠β.⑴求证:向量a＋向量b与向量a+向量b

(1)
(a+b).(a-b)
=(cosα+cosβ)(cosα-cosβ)+(sinα+sinβ)(sinα-sinβ)
=(cosα)^2- (cosβ)^2+(sinα)^2-(sinβ)^2
=0
=>a+b与a-b互相垂直
(2)
|ka+b|^2
=k^2|a|^2+|b|^2+2ka.b (1)
|a-kb|^2
=|a|^2+k^2|b|^2-2ka.b (2)
(1)=(2)
(k^2-1)|a|^2+(1-k^2)|b|^2 + 4ka.b=0
(k^2-1)+(1-k^2)+4k(cosαcosβ +sinαsinβ)=0
4kcos(β-α) =0
β-α = π/2