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Cosx+Cos2x+Cos3x+....+Cosnx=

cosx+cos2x+cos3x+....+cosnx =sin(x/2)*[ cosx+cos2x+cos3x+....+cosnx] / sin(x/2) ( 将sin(x/2) 移入方括号里并化简) = {sin[x(2n+1)/2] - sin(x/2) }/ [2sin(x/2)]

利用 e^(ix)=cosx+isinx; e^(ix)+e^(i2x)+e^(i3x)+……+e*(inx)=(cosx+cos2x+……+cosnx)+i(sinx+sin2x+……+sinnx) =[e^(inx+ix) -e^(ix)]/[e^(ix)-1]; 将最后一个等号右端分成实部和虚部(分母和分子同乘以 (cosx-1)-isinx),与等号左端实部和虚部...

cosx+cos2x+......+cosnx =1/2sin(x/2)*(cosx*2sin(x/2)+cos2x*2sin(x/2)+......+cosnx*2sin(x/2)) =1/2sin(x/2)*(sin(3x/2)-sin(x/2)+sin(5x/2)-sin(3x/2)+......+sin(n+1/2)x-sin(n-1/2)x) =1/2sin(x/2)*(sin(n+1/2)x-sin(x/2)) =1/2sin(x/2)*...

因为左边的和是实数,所以右边一定可以化为实数。在最后的化简,只需要提取exp(ix/2)和exp(inx/2)之类的就可以了。(个人喜好不同,会导致最后提取的是exp(-ix/2)和exp(-inx/2))

对于Sn=cosx+cos2x+cos3x+……+cosnx,有: 2sin(x/2)[cosx+cos2x+cos3x+……+cosnx ] =2sin(x/2)cosx+2sin(x/2)cos2x+2sin(x/2)cos3x+……+2sin(x/2)cosnx =sin(3x/2)-sin(x/2)+sin(5x/2)-sin(3x/2)+sin(7x/2)-sin(5x/2)+……+sin(x/2+nx)-sin(nx-x/2) ...

=1/2sin(x/2)*(cosx*2sin(x/2)+cos2x*2sin(x/2)+......+cosnx*2sin(x/2)) =1/2sin(x/2)*(sin(3x/2)-sin(x/2)+sin(5x/2...

解答:证明:∵2sinx2cosnx=sin(x2+nx)+sin(x2?nx).∴2sinx2(cosx+cos2x+…+cosnx)=(sin3x2?sinx2)+(sin5x2?3x2)+…+(sin1+2n2x?sin1?2n2x)=sin1+2n2x?sinx2=2cosn+12xsinn2x.∴cos+cos2x+…+cosnx=cosn+12x?sinn2xsinx2.

乘以2sinx, 积化和差就变成了 sin2x-0+sin3x-sinx+sin4x-sin2x+...+ sinnx-si(n-2)x+sin(n+1)x-sin(n-1)x =sin(n+1)x+sinnx-sinx 再除以2sinx,即为答案,[sin(n+1)x+sinnx-sinx]/2sinx

n(n+1)(2n+1)/12

没学洛必达==那就直接积化和差一步步展开吧,sinx~x这个知道伐?

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