Question

在△ABC中，sinA+sinB+sinC=0,cosA+cosB+cosC=0,求证：cos²A+cos²B+cos²C=3/2

Answer 1

cosa + cosb + cosc = 0
sina + sinb + sinc = 0

(cosa)^2 = (cosb + cosc)^2
= (cosb)^2 + (cosc)^2 + 2*cosb*cosc ............................(1)
(sina)^2 = (sinb + sinc)^2
= (sinb)^2 + (sinc)^2 + 2*sinb*sinc .................................(2)

(1) + (2)，得cos(b-c) = -1/2
同样可以得到：
cos(c-a) = -1/2
cos(a-b) = -1/2

(1) - (2)，得
cos2a = cos2b + cos2c + 2*cos(b+c)
= 2*cos(b+c)*cos(b-c) + 2*cos(b+c) ......... cos(b-c) = -1/2
= cos(b+c) .............................(3)
同样可以得到：
cos2b = cos(c+a) .............................(4)
cos2c = cos(a+b) .............................(5)

(cosa)^2 + (cosb)^2 + (cosc)^2
= (cos2a + cos2b + cos2c)/2 + 3/2
其中
A = cos2a + cos2b + cos2c
= [(cos2a + cos2b) + (cos2b + cos2c) + (cos2c +cos2a)]/2
= cos(a+b)*cos(a-b) + cos(b+c)*cos(b-c) + cos(c+a)*cos(c-a)
= -[cos(a+b) + cos(b+c) + cos(c+a)]/2
由(3)、(4)、(5)得到
A = cos(a+b) + cos(b+c) + cos(c+a)
所以，A = 0

cos²A+cos²B+cos²C=3/2

Answer 2

题可能有误。sinA+sinB+sinC=0????