高中三角函数题：化简cosx+cos2x+...+cosnx

Question

高中三角函数题：化简cosx+cos2x+...+cosnx

Answer 1

   cosx+cos2x+...+cosnx
=1/2[(cosx+cosnx)+(cos2x+cos(n-1)x)+...+(cosnx+cosx)]                              先乘以2后除以2
=[cos(n+1)x/2][cos((n-1)x/2)+cos(((n-3)x/2)+...+cos((n-(2n-1))x/2)               和差化积
=[cos(n+1)x/2/sin(x/2)]*[sin(x/2)*cos((n-1)x/2)+sin(x/2)*cos(((n-3)x/2)+...+sin(x/2)*cos((n-(2n-1))x/2)                                                                                            先乘以sin(x/2)后除以sin(x/2)
=1/2[cos(n+1)x/2/sin(x/2)][sin(nx/2)+sin((2-n)x/2)+sin((n-2)x/2)+sin((4-n)x/2)+...+sin((2-n)x/2)+sin(nx/2)]                                                                                                                  积化和差
={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)]                                                               整理化简
还有一种方法：

       经过和差化积公式可进一步化简,得到最终结果：{[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)]
       这两种方法用到了积化和差和和差化积公式,只要灵活掌握这两类公式,就好做了.
       以下为主要用到的几个公式：