若α、β、γ满足: sinα+sinβ+sinγ=0 (1) cocα+cosβ+cosγ=0 (2)则有 sinα=-(sinβ+sinγ)(3) cosα=-(cosβ+cosγ) (4) (3)的平方加上(4)的平方得: cos(β-γ)=-1/2 (5) 这是我们要证的第一个结论由(5)还可得:β-γ=2/3π+2kπ其中k∈Z。同理可证:γ-α=2/3π+2kπ,α-β=2/3π+2kπ、因此,在(1)、(2)条件下,有结论:α、β、γ依次相差2/3π+2kπ,(k∈Z)——这是要证的第二个结论。 If α, β, γ satisfy: sinα+sinβ+sinγ=0 (1) cocα+cosβ+cosγ=0 (2) then sinα=-(sinβ+sinγ) (3) cosα=-(cosβ+cosγ) ( 4) The square of (3) plus the square of (4): cos(β-γ)=-1/2 (5) This is the first conclusion we want to prove. (5) Also available: β- γ=2/3π+2kπ where k∈Z. Similarly, it can be proved that γ-α=2/3π+2kπ and α-β=2/3π+2kπ. Therefore, under conditions (1) and (2), it is concluded that α, β, and γ differ by 2 in order. 3π+2kπ, (k∈Z) - This is the second conclusion to prove.