cosx+cos2x+cos3x
@幸帝4623：cosx+cos2x+cos3x+....+cosnx= -
双陆17810733446…… 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 ...
@幸帝4623：高中三角函数题:化简cosx+cos2x+...+cosnx - 作业帮
双陆17810733446…… [答案] cosx+cos2x+...+cosnx=1/2[(cosx+cosnx)+(cos2x+cos(n-1)x)+...+(cosnx+cosx)] ...
@幸帝4623：三角恒等变换证明cosx+cos2x+…+cosnx=[cos(n+1/2)·sinnx/2]/sinx/2怎么证明? - 作业帮
双陆17810733446…… [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.
@幸帝4623：求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)] - 作业帮
双陆17810733446…… [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+...
@幸帝4623：求证cos x + cos2x +cos3x=cos2x(1+2cos x) -
双陆17810733446…… 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:
@幸帝4623：cosθ+cos2θ+······+cosnθ=? - 作业帮
双陆17810733446…… [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...
@幸帝4623：怎么化解cosx+cos2x+......+cosnx -
双陆17810733446…… 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
@幸帝4623：请数学高手解答 cosθ+cos2θ+······+cosnθ=? -
双陆17810733446…… (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)
@幸帝4623：!!高手来!!求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)] -
双陆17810733446…… 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...
@幸帝4623：数学三角函数化简3cosx+cos2x+cos3x -
双陆17810733446…… 求3cosx+cos2x+cos3x 我一个一个分解,你顺便记记公式,我还是喜欢用α,还是用α给你做好了 cos3α=cos(2α+α) =cos2α·cosα-sin2α·sinα =(cosα^2-sinα^2)cosα-2sinα^2·cosα =[cosα^2-(1-cosα^2)]cosα-2(1-cosα^2)cosα =(2cosα^2-1)cosα-2(1-...
双陆17810733446…… 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 ...
@幸帝4623：高中三角函数题:化简cosx+cos2x+...+cosnx - 作业帮
双陆17810733446…… [答案] cosx+cos2x+...+cosnx=1/2[(cosx+cosnx)+(cos2x+cos(n-1)x)+...+(cosnx+cosx)] ...
@幸帝4623：三角恒等变换证明cosx+cos2x+…+cosnx=[cos(n+1/2)·sinnx/2]/sinx/2怎么证明? - 作业帮
双陆17810733446…… [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.
@幸帝4623：求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)] - 作业帮
双陆17810733446…… [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+...
@幸帝4623：求证cos x + cos2x +cos3x=cos2x(1+2cos x) -
双陆17810733446…… 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:
@幸帝4623：cosθ+cos2θ+······+cosnθ=? - 作业帮
双陆17810733446…… [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...
@幸帝4623：怎么化解cosx+cos2x+......+cosnx -
双陆17810733446…… 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
@幸帝4623：请数学高手解答 cosθ+cos2θ+······+cosnθ=? -
双陆17810733446…… (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)
@幸帝4623：!!高手来!!求证:cosx+cos2x+...+cosnx={[cos(n+1)x/2]*[sin(n/2)x]}/[sin(x/2)] -
双陆17810733446…… 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...
@幸帝4623：数学三角函数化简3cosx+cos2x+cos3x -
双陆17810733446…… 求3cosx+cos2x+cos3x 我一个一个分解,你顺便记记公式,我还是喜欢用α,还是用α给你做好了 cos3α=cos(2α+α) =cos2α·cosα-sin2α·sinα =(cosα^2-sinα^2)cosα-2sinα^2·cosα =[cosα^2-(1-cosα^2)]cosα-2(1-cosα^2)cosα =(2cosα^2-1)cosα-2(1-...
