sina+sinb+sinc=0 cosa+cosb+cosc=0求证cos*2a+cos*2b+cos*2c=3|2

sinA+sinB=-sinC
cosA+cosB=-cosC
两式分别左右平方,后相加得
2+2(cosAcosB+sinAsinB)=1
所以
cosAcosB+sinAsinB=-1/2
cos(A-B)=cosAcosB+sinAsinB=-1/2

为(sinc)^2=(sina+sinb)^2=(sina)^2+2sina*sinb+(sinb)^2,
同理(cosc)^2=(cosa)^2+2cosa*cosb+(cosb)^2,
所以相加得1=1+2(cosa*cosb+sina*sinb)+1,
公式cos(a-b)=cosa*cosb+sina*sinb
∴所以相加得1=1+2cos(a-b)+1,
2cos(a-b)=-1
所以cos(a-b)=-1/2.

cosA+cosB=-cosC
sinA+sinB=-sinC
两个等式平方再相加得
2(cosAcosB+sinAsinB)=-1
cos(A-B)=-1/2
同理cos(B-C)=-1/2;cos(C-A)=-1/2;
cos(A-B)-cos(B-C)=-2sin((A-C)/2)sin((A+C)/2-B)=-1/2-(-1/2)=0
所以
(A-C)/2=kπ或(A+C)/2-B=kπ
若(A-C)/2=kπ
A=C+2kπ
cosB=-2cosC
sinB=-2sinC
两边平方相加得
1=4,矛盾
所以,(A+C)/2-B=kπ
由cos(B-C)=-1/2；cos(B-A)=-1/2
得A=B+2k‘π+2/3π或A=B+2k’π+4/3π
C=B+2k’‘π+2/3π或C=B+2k’‘π+4/3π
而再由(A+C)/2-B=kπ
A=B+2k‘π+2/3π且C=B+2k’‘π+4/3π
或C=B+2k‘π+2/3π且A=B+2k’‘π+4/3π
所以这道题的等式不成立
如B=10° A=130° C=250°
cosA+cosB+cosC=0,
sinA+sinB+sinC=0
sin(A+B+C)=sin(30°)=1/2 不等于sinA+sinB+sinC

这个人答了很详细.你自己去看看噢~

sin²a+cos²a=1
所以(-sinb-sinc)²+(-cosb-cosc)²=1
sin²b+cos²b+sin²c+cos²c+2sinbsinc+2cosbcosc=1
2+2(cosbcosc+sinbsinc)=1
cos(b-c)=cosbcosc+sinbsinc=-1/2
sin4θ+cos4θ
=(sin²θ+cos²θ)²-2sin²θcos²θ
=1-1/2*sin²(2θ)
=5/9
sin²(2θ)=8/9
2kπ+π

(sina)^2=(sinb+sinc)^2=(sinb)^2+2sinb*sinc+(sinc)^2,
(cosa)^2=(cosb+cosc)^2=(cosb)^2+2cosb*cosc+(cosc)^2,
两式相加得：1=1+2(cosb*cosc+sinb*sinc)+1,
所以 cos(b-c)=-1/2.

cos(B-C)=cosBcosC+sinBsinc又sinB+sinc=-sinA cosB+cosC=-cosA
所以同时平方sinB^2+sinc^2+2sinBsinc=sinA^2 cosB^2+cosc^2+2cosBcosC=cosA^2 两式相加 2（cosBcosC+sinBsinc）+2=1
cos(B-C)=cosBcosC+sinBsinc=-1/2

(1)已知sinA+sinB+sinC=0,cosA+cosB+cosC=0.求cos(B-C)的值.
sinA=-(sinB+sinC)
cosA=-(cosB+cosC)
sinA^2+cosA^2=1
带入整理可得cos(B-C)=-0.5
(2)在三角行ABC中,已知a^4+b^4+c^4=2c^2(a^2+b^2),则∠C=?
a^4+b^4+c^4=2c^2(a^2+b^2)
a^4+b^4+c^4-2c^2a^2-2c^2b^2=0
(a^2+b^2-c^2)^2=2a^2b^2
a^2+b^2-c^2=正负(根号2)ab
cosC=(a^2+b^2-c^2)/(2ab)=正负(根号2）/2
C=45度,或135度
(3)在三角行ABC中,已知∠A>∠B>∠C且∠A=2∠C,b=4,a+c=8,求a,c的长.
A=2C
sinB=sin(180-B)=sin(A+C)=sin3C
sinA=sin2C
由正弦定理得
b/sinB=(a+c)/(sinA+sinC)
4/sin3C=8/(sinA+sinC)
2sin3C=sin2C+sinC
之后用3倍2倍角及恒等式可得出8cos^2C-2cosC-3=0
所以cosC=3/4 (cosC=-1/2舍去)
sinC=√7/4
sinA=sin2C=3√7/8
a/c=sinA/sinc=3:2
所以a=24/5 c=16/5

sinb+sinc=-sina,cosb+cosc=-cosa
两个式两边同时平方且相见得
cosbcosc+sinbsinc=1
所以cos(b-c)=1
sinA+sinB=√2 /2……(1)
设cosA+cosB=S……(2)
(1)^2+(2)^2:1/2+1+2(cosAcosB+sinAsinB)=S^2+1
--->1/2+2cos(A-B)=S^2
--->S^2=1/2+2cos(A-B)
-1=

sinA+sinB=-sinC
cosA+cosB=-cosC
两式分别左右平方,后相加得
2+2(cosAcosB+sinAsinB)=1
所以
cosAcosB+sinAsinB=-1/2
cos(A-B)=cosAcosB+sinAsinB=-1/2
你把A换成B、B换成C就可以了

sinA+sinB=-sinC cosA+cosB=-cosC
两式平方相加得2+2cos(B-C)=1 cos(B-C)=-12
同理cos(C-A)=-12 cos(A-B)=-12
两式平方相减得cos2A+cos2B+cos(A+B)=cos2C
同理cos2B+cos2C+cos(B+C）=cos2A
cos2C+cos2A+cos(C+A)=cos2B
则-[cos(A+B)+cos(B+C)+cos(C+A)]=cos2A+cos2B+cos2C=12[2cos(A+B)cos(A-B)+
2cos(B+C)cos(B-C)+2cos(C+A)cos(C-A)]
=-12[cos(A+B)+cos(B+C)+cos(C+A)]
cos(A+B)+cos(B+C)+cos(C+A)=cos2A+cos2B+cos2C=0
Cos^2A+Cos^2B+Cos^2C=32+
12[cos2A+cos2B+cos2C]=32

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