cosx+cos2x+cos3x+cosx
答:cosxcos2xcos3x的不定积分为x/4+1/8sin2x+1/16sin4x+1/24sin6x+C。解:∫cosxcos2xcos3xdx =1/2∫cosx*(cos(3x+2x)+cos(3x-2x))dx =1/2∫cosx*(cos5x+cosx)dx =1/2∫cosxcos5xdx+1/2∫(cosx)^2dx =1/4∫(cos(5x+x)+cos(5x-x))dx+1/4∫(1+cos2x)dx =1/4∫1d...
答:(abc)'=a'bc+ab'c+abc'(cosxcos2xcos3x)'=-sinxcos2xcos3x-2cosxsin2xcos3x-3cosxcox2xsin3x 最后算出的结果是14吧
答:∫cosxcos2xcos3xdx=(1/2)∫(cosx+cos3x)cos3xdx=(1/2)∫cosxcos3xdx+(1/2)∫(cos3x)^2dx=(1/4)∫(cos2x+cos4x)dx+(1/4)∫(1+cos6x)dx=(1/4)∫dx+(1/4)∫cos2xdx+(1/4)∫cos4xdx...
答:cosx*cos2x*cos4x = 2 sinx*cosx*cos2x*cos4x / (2sinx)= sin2x * cos2x *cos4x /(2sinx) = sin8x / (8sinx)cos3x*cos5x =(1/2) ( cos8x +cos2x)原式= (1/16) (1/sinx) [ sin8x cos8x + sin8xcos2x ]= (1/32) (1/sinx) [ sin16x + sin10x +sin6x ]...
答:这个是倒数的四则运算和符合函数的导数 (abc)'=a'bc+ab'c+abc'(cosxcos2xcos3x)'=-sinxcos2xcos3x-2cosxsin2xcos3x-3cosxcox2xsin3x 最后算出的结果是14吧
答:∫cosxcos2xcos3xdx =(1/2)∫(cosx+cos3x)cos3xdx=(1/2)∫cosxcos3xdx+(1/2)∫(cos3x)^2dx =(1/4)∫(cos2x+cos4x)dx+(1/4)∫(1+cos6x)dx =(1/4)∫dx+(1/4)∫cos2xdx+(1/4)∫cos4xdx+(1/4)∫cos6xdx =(1/4)x+(1/8)sin2x...
答:用泰勒公式直接展开就可以了
答:可这样改:clf x=0:0.01:pi;y1=cos(x);y2=cos(2*x);y3=cos(3*x);y4=cos(4*x);a=plot(x,y1,'r');grid on hold on pause plot(x,y2,'g');pause plot(x,y3,'b')pause plot(x,y4,'y')legend('cosx','cos2x','cos3x','cos4x')hold off ...
答:cosx十cos2x十cos3x=0,求x;解:(cos3x+cosx)+cos2x=2cos2xcosx+cos2x=cos2x(2cosx+1)=0 由cos2x=0得2x=2kπ±(π/2),即有x₁,₂=kπ±(π/4),k∈Z;由2cosx+1=0,得cosx=-1/2;故x₃,4=2kπ±(2π/3),k∈Z;...
答:由cosx×cos2x×cos3x =cosx(2cos²x-1)(4cos³x-3cosx)=(2cos³-cosx)(4cos³x-3cosx)=8cos^6x-10cos^4x+3cos²x.
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亢砍13155985966:
高中三角函数题:化简cosx+cos2x+...+cosnx -
11056应祝
:[答案] cosx+cos2x+...+cosnx=1/2[(cosx+cosnx)+(cos2x+cos(n-1)x)+...+(cosnx+cosx)] ...
亢砍13155985966:
cosx+cos2x+cos3x+....+cosnx= -
11056应祝
: 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 ...
亢砍13155985966:
求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)] -
11056应祝
:[答案] cosx+cos2x+...+cosnx=1/2[(cosx+cosnx)+(cos2x+cos(n-1)x)+...+(cosnx+cosx)]=[cos(n+1)x/2][cos((n-1)x/2)+cos(((n-3)x/2)+...+cos((n-(2n-1))x/2)=[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)=1/2[cos(n+...
亢砍13155985966:
把cosx+cos2x+cos3x+cos4x化成积的形式. -
11056应祝
:[答案]分 析: 把cosx与cos4x看作一组,cos2x与cos3x看作一组进行和差化积.原式=(cosx+cos4x)+(cos2x+cos3x)=2coscos+2coscos=2cos·(cos+cos)=4coscosxcos.
亢砍13155985966:
求证cos x + cos2x +cos3x=cos2x(1+2cos x) -
11056应祝
: 证:cosx+cos2x+cos3x=(cosx+cos3x)+cos2x=2cos[(3x+x)/2]cos[(3x-x)/2] +cos2x=2cos2xcosx +cos2x=cos2x(1+2cosx) 用到的公式:
亢砍13155985966:
求cosα+cos2α+cos3α+.+cosnα的值 sinα+sin2α+sin3α+.+sinNα -
11056应祝
:[答案] cosx+cos2x+.cos(nx)=sin(nx/2)cos((n+1)x/2)/sin(x/2) sinx+sin2x+.sin(nx)=sin(nx/2)sin((n+1)x/2)/sin(x/2) 方法: cosx+cos2x+.cos(nx)+i[sinx+sin2x+.sin(nx)]=e^(ix)+e^(i2x)+...e^(inx) =e^(ix)*(1-e^(inx))/(1-e^(ix)) =e^(ix)*(sin(nx/2)/sin(x/2))e^(i(n-1)x/2) 整理分...
亢砍13155985966:
cosθ+cos2θ+······+cosnθ=? -
11056应祝
:[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...
亢砍13155985966:
化简:cosx+cos2x+cos3x+……cosnx=?化为最简式应该为什么呢? -
11056应祝
:[答案] 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
亢砍13155985966:
怎么化解cosx+cos2x+......+cosnx -
11056应祝
: 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
亢砍13155985966:
请数学高手解答 cosθ+cos2θ+······+cosnθ=? -
11056应祝
: (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+cos2x+...+cosnx)(cosx-1/2)=(1/2)(1+cos(n+1)x cosx+cos2x+...+cosnx=(1+cos(n+1)x)/(2cosx-1)