第3个回答 2008-01-15
Sin(a)+Sin(b)+Sin(c)=0 =>Sin(a)+Sin(b)=-Sin(c)...(1)
Cos(a)+Cos(b)+Cos(c)=0 =>Cos(a)+Cos(b)=-Cos(c)...(2)
(1)^2+(2)^2,化简,
=>Cos(a-b)=-1/2 =>|a-b|=2π/3(不考虑周期了)
由于对称性,可知|b-c|=2π/3,|a-c|=2π/3
只可能是a,b,c各相差2π/3把2π平分了
不妨设b=a+2π/3,c=a-2π/3
Cos^2(a)+Cos^2(b)+Cos^2(c)
=Cos^2(a)+Cos^2(a+2π/3)+Cos^2(a-2π/3)
化简,
=>原式=3/2