>>349
sin(b/2)*sin(a+bj) = (1/2){cos(a+b(j-1/2)) - cos(a+b(j+1/2))}
だから
左辺＝(1/2){cos(a+b(-1/2)) - cos(a+b(k-1/2))} = sin(a+(k-1)b/2)*sin(bk/2)