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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...

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))

我提供一个求均值的解答,方差是类似地。

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)*...

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) ...

dfh

cosx+cos2x =2cos²x-1+cosx =(2cosx-1)(cosx+1)

(I)证明:∵sin3x=?cos(3π2?3x)=?cos[3(π2?x)]=?[4cos3(π2?x)?3cos(π2?x)]=-(4sin3x-3sinx)=3sinx-4sin3x,故等式成立.(II)cos4x=cos(2?2x)=2cos22x-1=2(2cos2x-1)2-1=2(4cos4x-4cos2x+1)-1=8cos4x-8cos2x+1.(III)∵sin36°=cos...

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