高一数学暴难三角函数题,智商200以上才做得起?

令 t=Cos^2(a)+Cos^2(b)+Cos^2(c)
=3/2+1/2*(cos(2a)+cos(2b)+cos(2c)
Sin(a)+Sin(b)=-Sin(c) (1)
Cos(a)+Cos(b)=-Cos(c) (2)
(1)^2+(2)^2 => 2+2*Cos(a-b)=1
Cos(a-b)=-1/2
同理
Cos(b-c)=-1/2
Cos(c-a)=-1/2
Cos(2a)+cos(2b)=2*cos(a+b)*cos(a-b)=-cos(a+b)
Cos(2b)+cos(2c)=-cos(b+c)
Cos(2c)+cos(2a)=-cos(c+a)
即 2*t=3+(-1/2)*[cos(a+b)+cos(b+c)+cos(c+a)] (3)
(1)^2-(2)^2 =>
2*Cos^2(a)+2*Cos(a+b)+2*Cos^2(b)=2*Cos^(c)+1
2*Cos^2(b)+2*Cos(b+c)+2*Cos^2(c)=2*Cos^(a)+1
2*Cos^2(c)+2*Cos(c+a)+2*Cos^2(a)=2*Cos^(b)+1
2*t+2*[cos(a+b)+cos(b+c)+cos(c+a)]=3 (4)
比较(3),(4),易得 cos(a+b)+cos(b+c)+cos(c+a)=0
得 t=3/2.

0

用二倍角即可

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