由已知向量a=(sino,1),b=(1,coso),- π/2＜o＜π/2,（1）若a⊥b,那么,向量a ·b =sin o ＋cos o =0 ,sin o =-cos o ,因为sin ²o ＋cos ²o =2sin ²o =1解得sin o =-√2/2,cos o =√2/2即o =-π/4或sin o =√2/2,cos o =-√2/2不和题意舍去
(2)（a+b）的平方＝(1＋sin o )²＋(1＋cos o)²=3＋2(sin o ＋cos o )=3＋2√2sin (o＋π/4)最大值为3 ＋2√2在o＝π/4处取得,故绝对值的最大值为√(3 ＋2√2)＝1＋√2