cosx+cos2x+cos3x+cosx
@仉怜3591：cosx+cos2x+cos3x+....+cosnx= -
荀卿15147003561…… 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 ...
@仉怜3591：高中三角函数题:化简cosx+cos2x+...+cosnx - 作业帮
荀卿15147003561…… [答案] cosx+cos2x+...+cosnx=1/2[(cosx+cosnx)+(cos2x+cos(n-1)x)+...+(cosnx+cosx)] ...
@仉怜3591：求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)] - 作业帮
荀卿15147003561…… [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+...
@仉怜3591：求证cos x + cos2x +cos3x=cos2x(1+2cos x) -
荀卿15147003561…… 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:
@仉怜3591：三角恒等变换证明cosx+cos2x+…+cosnx=[cos(n+1/2)·sinnx/2]/sinx/2怎么证明? - 作业帮
荀卿15147003561…… [答案] 说个思路啊,用2sinx/2分别乘以等式左边各项,然后用积化和差公式,最后发现有好多项消去了.最后就得你要的东西了.
@仉怜3591：求证:cosx+cos2x+…+cosnx=cosn+12x•sinn2xsinx2. - 作业帮
荀卿15147003561…… [答案] 证明:∵2sin x 2cosnx=sin( x 2+nx)+sin( x 2−nx). ∴2sin x 2(cosx+cos2x+…+cosnx)=(sin 3x 2−sin x 2)+(sin 5x 2− 3x 2)+…+(sin 1+2n 2x−sin 1−2n 2x) =sin 1+2n 2x−sin x 2 =2cos n+1 2xsin n 2x. ∴cos+cos2x+…+cosnx= cosn+12x•sinn2x sinx2.
@仉怜3591：怎么化解cosx+cos2x+......+cosnx -
荀卿15147003561…… 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
@仉怜3591：化简:cosx+cos2x+cos3x+……cosnx=?化为最简式应该为什么呢? - 作业帮
荀卿15147003561…… [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
@仉怜3591：请数学高手解答 cosθ+cos2θ+······+cosnθ=? -
荀卿15147003561…… (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)
@仉怜3591：!!高手来!!求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)] -
荀卿15147003561…… 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 我都快吓晕了....然后...
荀卿15147003561…… 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 ...
@仉怜3591：高中三角函数题:化简cosx+cos2x+...+cosnx - 作业帮
荀卿15147003561…… [答案] cosx+cos2x+...+cosnx=1/2[(cosx+cosnx)+(cos2x+cos(n-1)x)+...+(cosnx+cosx)] ...
@仉怜3591：求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)] - 作业帮
荀卿15147003561…… [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+...
@仉怜3591：求证cos x + cos2x +cos3x=cos2x(1+2cos x) -
荀卿15147003561…… 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:
@仉怜3591：三角恒等变换证明cosx+cos2x+…+cosnx=[cos(n+1/2)·sinnx/2]/sinx/2怎么证明? - 作业帮
荀卿15147003561…… [答案] 说个思路啊,用2sinx/2分别乘以等式左边各项,然后用积化和差公式,最后发现有好多项消去了.最后就得你要的东西了.
@仉怜3591：求证:cosx+cos2x+…+cosnx=cosn+12x•sinn2xsinx2. - 作业帮
荀卿15147003561…… [答案] 证明:∵2sin x 2cosnx=sin( x 2+nx)+sin( x 2−nx). ∴2sin x 2(cosx+cos2x+…+cosnx)=(sin 3x 2−sin x 2)+(sin 5x 2− 3x 2)+…+(sin 1+2n 2x−sin 1−2n 2x) =sin 1+2n 2x−sin x 2 =2cos n+1 2xsin n 2x. ∴cos+cos2x+…+cosnx= cosn+12x•sinn2x sinx2.
@仉怜3591：怎么化解cosx+cos2x+......+cosnx -
荀卿15147003561…… 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
@仉怜3591：化简:cosx+cos2x+cos3x+……cosnx=?化为最简式应该为什么呢? - 作业帮
荀卿15147003561…… [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
@仉怜3591：请数学高手解答 cosθ+cos2θ+······+cosnθ=? -
荀卿15147003561…… (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)
@仉怜3591：!!高手来!!求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)] -
荀卿15147003561…… 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 我都快吓晕了....然后...
