设α、β、γ∈R,且sinα+sinγ=sinβ,cosα+cosγ=cosβ,则α-β()-π/3 π/6 -π/3或π/3 π/3

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设α、β、γ∈R,且sinα+sinγ=sinβ,cosα+cosγ=cosβ,则α-β()-π/3 π/6 -π/3或π/3 π/3

设α、β、γ∈R,且sinα+sinγ=sinβ,cosα+cosγ=cosβ,则α-β()-π/3 π/6 -π/3或π/3 π/3
设α、β、γ∈R,且sinα+sinγ=sinβ,cosα+cosγ=cosβ,则α-β()
-π/3 π/6 -π/3或π/3 π/3

设α、β、γ∈R,且sinα+sinγ=sinβ,cosα+cosγ=cosβ,则α-β()-π/3 π/6 -π/3或π/3 π/3
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 ...)