Question

若点O为三角形ABC的外心，且2向量OA+2向量OB+3向量OC=0，C为三角形ABC的内角，cos2C=多少

Answer 1

O为三角形ABC的外心
=>|OA|=|OB|=|OC|
2OA+2OB+3OC=0
|2OA+2OB|=|3OC|
(2OA+2OB)(2OA+2OB)=9|OC|^2
4|OA|^2+4|OB|^2+8|OA||OB|cos∠AOB= 9|OC|^2
8+8cos∠AOB=9
cos∠AOB = 1/8
similarly
|2OA+3OC|=|2OB|
=> cos∠AOC= -3/4
|2OB+3OC|=|2OA|
=>cos∠BOC= -3/4
∠C = (π-∠BOC)/2 + (π-∠AOC)/2
= π- (∠BOC+∠AOC)/2
2C = 2π- (∠BOC+∠AOC)
cos2C= cos(∠BOC+∠AOC)
=(cos∠BOCcos∠AOC-sin∠BOCsin∠AOC)
= (-3/4)( -3/4) - ( √7/4) √7/4)
= 1/8