Question

高一数学题，求过程，有好评

Answer 1

解：1、(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)
2、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

Answer 2

SinA + SinB = -Sinc (1)
CosA + CosB = -CosC (2)
平方相加(1)(2)
2+2Cos(A-B) = 1 (3)
平方（2）-（1）
Cos2A + Cos2B + 2Cos(A+B) = Cos2C

2Cos(A+B)Cos(A-B) + 2Cos(A+B) = Cos2C
Cos(A+B){2Cos(A-B) + 2} = Cos2C
Cos(A+B) = Cos2C 利用（3）
因为SinA+SinB+SinC = 0
平方Sin2A+Sin2B+Sin2C = -2∑SinASinB

因为 Cos(A+B) = Cos2C
同理 Cos(B+C) = Cos2A
Cos(C+A) = Cos2B
相加∑CosACosB - ∑SinASinB = ∑Cos2A

Cos(A-B) = -1/2 利用（3）
同理Cos(B-C) = -1/2
Cos(C-A) = -1/2

相加∑CosACosB + ∑SinASinB = -3/2
所以 -2∑SinASinB = ∑Cos2A + 3/2
从（4）可得 Sin2A+Sin2B+Sin2C = -2∑SinASinB = ∑Cos2A + 3/2

Cos2A+Cos2B+Cos2C = 2Cos(A+B)Cos(A-B) + Cos2C = -Cos(A+B) + Cos2C = 0
又因为 Cos(A+B) = Cos2C
Cos(A-B) = -1/2
因此 ∑Cos2A = 0.
因此SinA平方+SinB平方+SinC平方 = 3/2.