题目

已知A,B,C三点的坐标分别为A(3,0),B(0,3),C(cosa,sina),a属于(兀/2,3兀/2),
若向量AC.向量BC=-1,求（2sin^2a+sin2a)(tana/2+1/tan(a/2)的值

提问时间：2021-01-03

答案

向量AC = (cosa - 3,sina)
向量BC = (cosa,sina - 3)
向量AC·向量BC
= cos²a - cosa + sin²a - 3sina
= 1 - 3(sina+cosa)
= - 1
∴ sina + cosa = 2/3
tan(a/2) + 1/tan(a/2)
= sin(a/2) / cos(a/2) + cos(a/2) / sin(a/2)
= [ sin²(a/2) + cos²(a/2) ] / [ sin(a/2) cos(a/2) ]
= 1 / ( 1/2 sina)
= 2 / sina
原式 = (2sin²a + sin2a ) * 2/sina
= (2sin²a + 2sinacosa) *2/sina
= 4sina + 4cosa
= 8/3