Question

高中数学导学导练满足三角函数等于问题

只有图片一张
最好解析一下

Answer 1

cosa + cosb + cosc = 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

(cosa)^2 + (cosb)^2 + (cosc)^2 = 3/2
参考资料：http://zhidao.baidu.com/link?url=Aic0atRINYsBW2stigzyJ8o57-NtmN3byKUOS81E_4wjUCg0yxOLydTGPrPvc8XZFHj6eiY3x1EfOKvAzJBnN_

Answer 2

建立以O为圆心的单位圆 设A,B C的终边与单位圆分别交于 M,N,P 它们的横坐标分别为 cosA ,cosB,
cosC 纵坐标分别为 sinA,sinB,sinC 因为sinA+sinB+sinC=0① cosA +cosB+cosC=0 ②
所以 向量OM+向量ON+向量OP=0 即M N P在单位圆的圆周的三等分点上
所以cos(A-B)=cos(±120°)=-1/2 同理cos（A-C)=-1/2 cos(B-C)=-1/2
②²-①² 得 cos2A+cos2B+cos2C+2[cos(A+B)+cos(A+C)+cos(B+C)]=0③
∵cos2A+cos2B=2cos(A+B)cos(A-C)=-cos(A+B)④
同理 cos2A+cos2C=-cos(A+C)⑤ cos2B+cos2C=-cos(B+C)⑥
④+⑤+⑥整理得 cos(A+B）+cos(A+C）+cos(B+C)=-2（cos2A+cos2B+cos2C)⑦
⑦代入③整理得cos2A+cos2B+cos2C=0 利用倍角公式=>cos²A+cos²B+cos²C=3/2