记T=sina+sin(a+b)+...+sin(a+nb)
两边同时乘以sin(b/2):
Tsin(b/2)=sinasinb/2+sin(a+b)sinb/2+...sin(a+nb)sinb/2
再用积化和差公式：sinasinb=[cos(a-b)-cos(a+b)]/2
得Tsin(b/2)=[cos(a-b/2)-cos(a+b/2)+cos(a+b/2)-cos(a+3b/2)+cos(a+3b/2)-cos(a+5b/2)+...
+cos(a+nb-b/2)-cos(a+nb+b/2)]/2
正负项抵消,得
Tsin(b/2)=[cos(a-b/2)-cos(a+nb+b/2)]/2,再用和差化积公式：
Tsin(b/2)=-sin(a+nb/2)sin(-nb/2-b/2)
故有T=sin(a+nb/2)sin(nb/2+b/2)/sin(b/2)
得证.