设α、β、γ∈R,且sinα+sinγ=sinβ,cosα+cosγ=cosβ,则α-β()

sinγ=sinβ - sinα,cosγ=cosβ - cosα(sinβ - sinα)^2 +(cosβ - cosα)^2 = (sinγ)^2 + (cosγ)^2 = 1 (1)同时展开 (sinβ - sinα)^2 +(cosβ - cosα)^2 =2 - 2 (sinβ sinα + cosβ cosα）= 2 - 2cos(β -α) (2)结合（1）（2）两式得到 2 - 2cos(β -α) = 1 得到 cos(β -α) = 1/2,β -α = π/3 + 2kπ 或 -π/3 + 2kπ k∈(。。。,-1,0,1,2 。。。)