cosxcos2xcos3x =1/2(cosx+cos3x)cos3x =1/4(cos2x+cos4x+cos6x+1)=1/4{[1-2x^2+o1(x^2)]+[1-8x^2+o2(x^2)]+[1-18x^2+o3(x^2)]+1} =1-7x^2+o(x^2)∴a=7，n=2
化简:cosx+cos2x+cos3x+……cosnx=?化为最简式应该为什么呢? 相关知识点: 试题来源: 解析 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 ...
1-cosxcos2xcos3x = 1 - [1-(1/2)x^2-(1/2)(2x)^2-(1/2)(3x)^2 + o(x^2)]= (1/2)(1+4+9)x^2 + o(x^2) = 7x^2 + o(x^2)
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-cosα^2)cosα （说明:cos2α可以一步导出cos2α=2cosα^2-1）=cosα[2cosα^2-1-2(1-...
0.5+cosx+cos2x+cos3x………cosnx,把这个式子化简成分子和分母的形式 答案 把原式乘以sinx后再除以sinx原式={1/2(sinx)+1/2(sin2x-sin0)+1/2(sin3x-sinx)+1/2(sin4x-sin2x)+………+1/2[sin(n+1)x-sin(n-1)x]}/sinx=1/2[sin(nx)+sin(n+1)x]/sinx=sin{[(2n+1)/2]x}cos(x/2...
希望你能看懂 张宇的1000题里用过这个
sinnxsin(x 2)=1 2(cos(nx+x 2)-cos(nx-x 2))=1 2(cos2n+1 2-cos2n-1 2)以上各式相加得sinx 2S=1 2(cos2n+1 2-cosl X),所以原式=1 1 cos(n+-)x-cos-x 2 2 2sin 2=n+l nx sin xsin 2 2 sin- 2(2)设S=cosx+cos2x+cos3x+…+cosnx,将式子两边乘以sinx 2,cosxsinx 2...
= 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 ]
【题目】如何把cosx+cos2x化简为 2(cos3*/2)(cos1*12)?(其中 x=3*/2-1*/2) 答案 【解析】cosx+cos2x=cosx+cosx^2-sinx^2=cosx+2cosx^2-1相关推荐 1如何把cosx+cos2x化简为2(cos3x/2)(cos1x/2)?(其中x=3x/2-1x/2) 2【题目】如何把cosx+cos2x化简为 2(cos3*/2)(cos1*12)...