高中三角函数题:化简cosx+cos2x+...+cosnx
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高中三角函数题:化简cosx+cos2x+...+cosnx
![高中三角函数题:化简cosx+cos2x+...+cosnx](/uploads/image/z/1849777-25-7.jpg?t=%E9%AB%98%E4%B8%AD%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E9%A2%98%EF%BC%9A%E5%8C%96%E7%AE%80cosx%2Bcos2x%2B...%2Bcosnx)
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:
![](http://img.wesiedu.com/upload/2/62/2623b53589907fec6ca9b85514f77a66.jpg)
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:
![](http://img.wesiedu.com/upload/a/3a/a3a421b0feec0ef798f510a41f5903f1.jpg)
![](http://img.wesiedu.com/upload/3/e3/3e35d5513f3e5a4fe63af609d68d0ba9.jpg)
=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:
![](http://img.wesiedu.com/upload/2/62/2623b53589907fec6ca9b85514f77a66.jpg)
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:
![](http://img.wesiedu.com/upload/a/3a/a3a421b0feec0ef798f510a41f5903f1.jpg)
![](http://img.wesiedu.com/upload/3/e3/3e35d5513f3e5a4fe63af609d68d0ba9.jpg)
高中三角函数题:化简cosx+cos2x+...+cosnx
cosx+cos2x+cos3x+.+cosnx=
怎么化解cosx+cos2x+.+cosnx
化简:cosx+cos2x+cos3x+……cosnx=?
求和Sn=cosx+cos2x+cos3x+……+cosnx
COSX+COS2X+COS3X+COS4X+COS5X+COS6X+...+COSNX=1/2|{SIN(N+1/2)
求证:cosx+cos2x+…+cosnx=cosn+12x•sinn2xsinx2
lim[(1-cosx*cos2x****cosnx)/x^2]在x趋于0时
C(N,1)COSX+C(N,2)COS2X+-----+C(N,N)COSNX
化简(cos2x/sinx+cosx)-(cos2x/sinx-cosx)
0.5+cosx+cos2x+cos3x…………cosnx,把这个式子化简成分子和分母的形式
求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2