cosnx的累加求和

程萱该1656怎么化解cosx+cos2x+......+cosnx -
葛甘雪19817231572 ______ 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)*(2*sinnx*cos(n+1)x)=(sinnx*cos(n+1)x)/sin(x/2) 或用,[sin(n+1/2)x/sin(x/2)]/2-1/2

程萱该1656cosx+cos2x+cos3x+....+cosnx= -
葛甘雪19817231572 ______ 具体回答如下: cosx+cos2x+cos3x+....+cosnx =sin(x/2)*[ cosx+cos2x+cos3x+....+cosnx] / sin(x/2) = 【sin[x(2n+1)/2] - sin(x/2) 】/ [2sin(x/2)] 和角公式: sin ( α ± β ) = sinα · cosβ ± cosα · sinβ sin ( α + β + γ ) = sinα · cosβ · cosγ + cosα · sinβ ...

程萱该1656cos(n+1)x=cos(nx)cosx - sin(nx)sinx这个式子是怎推出来的? -
葛甘雪19817231572 ______[答案] 根据公式cos(x+y)=cosxcosy-sinxsiny带入即可.

程萱该1656cosθ+cos2θ+······+cosnθ=? -
葛甘雪19817231572 ______[答案] (cosx+cos2x+cos3x+...+cosnx)cosx=(cosxcosx+cosxcos2x+...+cosxcosnx)=(1/2)(cos2x+1)+(1/2)(cos3x+cosx)+(1/2)(cos4x+cos3x)+...+(1/2)(cosn+1)x+cosnx)=(1/2)(1+cos(n+1)x) +(1/2)(cosx+cos2x+...+cosnx)(cosx+c...

程萱该1656求积分 cos^n(x) *cos(nx)求此式子积分(cos(x))^n * cos(nx) -
葛甘雪19817231572 ______[答案] =1/(n+1)[sin(nx)*cos^n(x)]

程萱该1656sin1x+sin2x+sin3x+sin4x+…+sinnx…一直加下去等于多少? -
葛甘雪19817231572 ______[答案] sinX乘以sin(X/2)=1/2(cos(x+x/2)-cos(x-x/2))sin2X乘以sin(X/2)=1/2(cos(2x+x/2)-cos(2x-x/2))sin3X乘以sin(X/2)=1/2(cos(3x+x/2)-cos(3x-x/2)).sinNX乘以1/2sin(X/2)=1/2(cos(Nx+x/2)-cos(Nx-x/2))发现上...

程萱该1656如何求sinx+sin2xsin3x+...+sinnx -
葛甘雪19817231572 ______ 用复数比较方便 e^(xi)=cosx+isinx e^(2xi)=cos2x+isin2x ... e^(nxi)=cosnx+isinnx 相加得 e^(xi)+e^(2xi)+...+e^(nxi)=(cosx+cos2x+...+cosnx)+i(sinx+sin2x+...+sinnx) e^(xi)+e^(2xi)+...+e^(nxi), 用等比数列求和 =e^(xi) [1-e^(nxi)]/(1-e^(xi)) =Cos[(1 + n) x/...

程萱该1656sinx+sin2x+sin3x+.+sinnx=?求和.数列求和里的一道题. -
葛甘雪19817231572 ______[答案] 设S=sinx+sin2x+sin3x+……+sinnx 两边同乘以2sin(x/2)(x≠2kπ,k∈Z) 得2sin(x/2)S=cos(x/2)-cos[(2n+1)x/2]=2sin(nx/2)sin[(n+1)x/2] 所以S=sin(nx/2)sin[(n+1)x/2]÷sin(x/2)

程萱该1656cos((n+1)x)=2cos(x)cos(nx) - cos((n - 1)x) 怎么证明? -
葛甘雪19817231572 ______[答案] 利用cosa+cosb=2cos[(a+b)/2]cos[(a-b)/2],得: cos[(n+1)x]+cos[(n-1)x]=2cos{[(n+1)x+(n-1)x]/2}cos{[(n+1)x-(n-1)x]/x}=2cos(x)cos(nx) 则:cos[(n+1)x]=2cos(x)cos(nx)-cos[(n-1)x]