求证sin(a/2)(sina+sin2a++sin3a+..+sinna)=sin(na/2)*sin[(n+1)a/2]

Question

求证sin(a/2)(sina+sin2a++sin3a+..+sinna)=sin(na/2)*sin[(n+1)a/2]

Answer 1

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...