求用C++语言实现cos(α1+α2+α3+……+αn) = 和sin(α1+α2+α3+……+αn) =的分解cos(α+β) = cosαcosβ- sinαsinβ,sin(α+β) = sinαcosβ+ cosαsinβ cos(α+β+γ) = cos(α+(β+γ)) = cosαcos(β+γ) - sinαsin(β+γ) =cosαcosβcosγ-c
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![求用C++语言实现cos(α1+α2+α3+……+αn) = 和sin(α1+α2+α3+……+αn) =的分解cos(α+β) = cosαcosβ- sinαsinβ,sin(α+β) = sinαcosβ+ cosαsinβ cos(α+β+γ) = cos(α+(β+γ)) = cosαcos(β+γ) - sinαsin(β+γ) =cosαcosβcosγ-c](/uploads/image/z/8702546-50-6.jpg?t=%E6%B1%82%E7%94%A8C%2B%2B%E8%AF%AD%E8%A8%80%E5%AE%9E%E7%8E%B0cos%28%CE%B11%2B%CE%B12%2B%CE%B13%2B%E2%80%A6%E2%80%A6%2B%CE%B1n%29+%3D+%E5%92%8Csin%28%CE%B11%2B%CE%B12%2B%CE%B13%2B%E2%80%A6%E2%80%A6%2B%CE%B1n%29+%3D%E7%9A%84%E5%88%86%E8%A7%A3cos%28%CE%B1%2B%CE%B2%29+%3D+cos%CE%B1cos%CE%B2-+sin%CE%B1sin%CE%B2%2Csin%28%CE%B1%2B%CE%B2%29+%3D+sin%CE%B1cos%CE%B2%2B+cos%CE%B1sin%CE%B2+cos%28%CE%B1%2B%CE%B2%2B%CE%B3%29+%3D+cos%28%CE%B1%2B%28%CE%B2%2B%CE%B3%29%29+%3D+cos%CE%B1cos%28%CE%B2%2B%CE%B3%29+-+sin%CE%B1sin%28%CE%B2%2B%CE%B3%29+%3Dcos%CE%B1cos%CE%B2cos%CE%B3-c)
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