求证sin(a/2)(sina+sin2a++sin3a+..+sinna)=sin(na/2)*sin[(n+1)a/2]
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![求证sin(a/2)(sina+sin2a++sin3a+..+sinna)=sin(na/2)*sin[(n+1)a/2]](/uploads/image/z/13162898-2-8.jpg?t=%E6%B1%82%E8%AF%81sin%28a%2F2%29%28sina%2Bsin2a%2B%2Bsin3a%2B..%2Bsinna%29%3Dsin%28na%2F2%29%2Asin%5B%28n%2B1%29a%2F2%5D)
求证sin(a/2)(sina+sin2a++sin3a+..+sinna)=sin(na/2)*sin[(n+1)a/2]
求证sin(a/2)(sina+sin2a++sin3a+..+sinna)=sin(na/2)*sin[(n+1)a/2]
求证sin(a/2)(sina+sin2a++sin3a+..+sinna)=sin(na/2)*sin[(n+1)a/2]
证明:
由积化和差公式:sinαsinβ=-1/2[cos(α+β)-cos(α-β)]得
sin(a/2)*sina = 1/2 * cos(3*a/2) - 1/2 * cos(a/2)
sin(a/2)*sin2*a = 1/2 * cos(5*a/2) - 1/2 * cos(3*a/2)
sin(a/2)*sin3*a = 1/2 * cos(7*a/2) - 1/2 * cos(5*a/2)
.
sin(a/2)*sinn*a = 1/2 * cos[(2*n-1)*a/2] - 1/2 * cos[(2*n-1)*a/2]
所以左边 = 1/2*cos[(2*n+1)*a/2] - 1/2*cos(a/2)
再由和差化积公式 cosθ-cosφ=-2sin[(θ+φ)/2]sin[(θ-φ)/2]得
左边 = 1/2*cos[(2*n+1)*a/2] - 1/2*cos(a/2)
= sin(na/2)*sin[(n+1)a/2]