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已知F(x)=CosxCos2x……Cosnx,求F"(0)

先用2^n·sinx乘表达式的两边,求得f(x)=sin2nx/(2^n·sinx),然后再根据商的求导法则和复合函数求导法则,很容易求出f''(x),进而求出f''(0).

解答:证明:我们知道2sinx2(12+cosx+cos2x+…+cosnx)=sinx2+(sin32x?sinx2)+…+[sin(n+12)x?sin(n?12)x]=sin(n+12)x,于是12+cosx+cos2x+…+cosnx=sin(n+12)x2sinx2,由此可得我们知道π2=∫π0sin(n+12)x2sinx2dx=∫π0sin(2n+1)u2sinu2du=∫π0sin(2n...

希望你学过复数的三角形式... 设z=cosx+isinx 由棣美弗定理 z^n=cosnx+isinnx 则上式左边即为 z+z^2+z^3+...+z^n的实部 又z+z^2+...+z^n=z(1-z^n)/(1-z) =(cosx+isinx)(1-cosnx-isinnx)(1-cosx+isinx)/[(1-cosx)^2+sin^2x] 确实很冗长 我都快吓...

原式=lim(1-cos2x)/x^2 =lim(2sin^x)/x^2 =2 如有疑问请追问 满意请采纳 如有其它问题请采纳此题后点求助, 答题不易,望合作O(∩_∩)O~

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)*...

dfh

当x→0时,cosx=cos2x=……=cosnx→1 原式=0

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.

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) ...

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