2sin(b/2)cos(a+kb)=sin[a+(k+1/2)b]-sin[a+(k-1/2)b],∑2sin(b/2)cos(a+kb)=sin[a+(n+1/2)b]-sin[a-b/2]=2cos(a+nb/2)sin[(n+1)b/2],所以(1)式成立。同理可知(2)式成立。
