1cosxcos2xcos3x与axn
@沙馥800：(1 - cosxcos2xcos3x)/(1 - cosx)当x趋近于0时的极限 -
严光18588658759…… 由三角积化和差公式 cosxcos2xcos3x =(1/2)(cosx+cos3x)xos3x =(1/4)cos2x+(1/4)cos4x+1/4+(1/4)cos6x 原极限化为(x->0) (1-(1/4)cos2x-(1/4)cos4x-1/4-(1/4)cos6x)/(1-cosx) x->0 1-cosx~(1/2)x^2 上式=(1-(1/4)cos2x-(1/4)cos4x-1/4-(1/4...
@沙馥800：cosxcos2xcos3x的导数 -
严光18588658759…… 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
@沙馥800：三角函数规律问题谁会求这个式子啊!cosxcos2xcos4xcos8x=?还有一个题 已知2sinx+3siny=a ,2cosx - 3cosy=b,求cos(x+y)? - 作业帮
严光18588658759…… [答案] (sinx*cosxcos2xcos4xcos8x)/sinx sin2xcos2xcos4xcos8x/2sinx sin4xcos4xcos8x/4sinx sin8x*cos8x/8sinx sin16x/16sinx
@沙馥800：计算cosxcos2xcos3x….cosnx
严光18588658759…… cosx+cos2x+cos3x+......+cosnx =1/(2sin(x/2))*[2cosxsin(x/2)+2cos2xsin(x/2)+......+2cosnxsin(x/2)] 括号中的数列的和等于“ [sin(3x/2)-sin(x/2)]+[sin(5x/2)-sin(3x/2)]+[sin(7x/2)-sin(5x/2)]+...... +{sin[(2n+1)x/2]-sin[(2n-1)x/2]} =sin[(2n+1)x/2]-sin(x/2) =2cos(n+1)xsin(nx/2) 所以,原式=cos(n+1)xsinnx/sin(x/2).
@沙馥800：∫cosxcos2xcos3xdx -
严光18588658759…… cosxcos2xcos3x = cos2x * cosxcos3x = cos2x * (1/2)[cos(x + 3x) + cos(x - 3x)] = cos2x * (1/2)[cos4x + cos2x] = (1/2)cos2xcos4x + (1/2)cos²2x = (1/2)(1/2)[cos(2x + 4x) + cos(2x - 4x)] + (1/2)(1/2)(1 + cos4x) = (1/4)cos6x + (1/4)cos2x + 1/4 + (1/4)cos4...
@沙馥800：1.(cosx)^3={(cos2x+1)*(cosx)}/2 是怎么化简得来的?要步骤! -
严光18588658759…… 1,用了个二倍角公式cos2x=2(cosx)^2-1 (cosx)^2=(cos2x+1)/2 2,应该是cos2xcosx=(cos3x+cosx)/2 这个采用了积化和差公式cos3x=cos(2x+x)=cos2xcosx- cosx=cos(2x-x)=cos2xcosx+sin2xsinx 两式相加得到2cos2xcosx=cos3x+cosx
@沙馥800：cosx+cos2x+cos3x+....+cosnx= -
严光18588658759…… 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 ...
@沙馥800：cosxcos2x=1/2(cosx+cos3x) 这个没看懂 求大神解释.. -
严光18588658759…… 积化和差公式现在的教材中已经删除了,可以用下面的替代: 思路分析: 找到角 x与3x 的平均值;2x 让平均值 2x 出场,以平均值为主线;更改原来的角的样式过渡到左边的角 x,及2x, 思路启蒙于等差数列; cosx+cos3x=cos(2x-x)+cos(2x+x) =[cos2xcosx+sin2xsinx]+[cos2xcosx-sin2xsinx] =2cos2xcosx 两边同除以 2得: cosxcos2x=(1/2)(cosx+cos3x)
@沙馥800：∫cosxcos2xcos3xdx - 作业帮
严光18588658759…… [答案] cosxcos2xcos3x = cos2x * cosxcos3x = cos2x * (1/2)[cos(x + 3x) + cos(x - 3x)] = cos2x * (1/2)[cos4x + cos2x] = (1/2)cos2xcos4x + (1/2)cos²2x = (1/2)(1/2)[cos(2x + 4x) + cos(2x - 4x)] + (1/2)(1/2)(1 + cos4x) = (1/4)cos6x + (1/4)cos2x + 1/4 + (1/4)cos4x ∫ ...
@沙馥800：当X趋向0时,1 - cos(X)cos(2X)cos(3X)对于X的无穷小的阶等于?即问与X的几次方是同阶无穷小. - 作业帮
严光18588658759…… [答案] A= [1-cosxcos2xcos3x]/x^2用洛必达法则分子分母求导 A1= [sinx cos2xcos3x + 2cosx sin2xcos3x+3 cosxcos2xsin3x ]/(2x)分子分母求导 A2= [cosx cos2xcos3x -2sinx sin2xcos3x -3sinx cos2xsin3x - 2sinx sin2xcos3...
严光18588658759…… 由三角积化和差公式 cosxcos2xcos3x =(1/2)(cosx+cos3x)xos3x =(1/4)cos2x+(1/4)cos4x+1/4+(1/4)cos6x 原极限化为(x->0) (1-(1/4)cos2x-(1/4)cos4x-1/4-(1/4)cos6x)/(1-cosx) x->0 1-cosx~(1/2)x^2 上式=(1-(1/4)cos2x-(1/4)cos4x-1/4-(1/4...
@沙馥800：cosxcos2xcos3x的导数 -
严光18588658759…… 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
@沙馥800：三角函数规律问题谁会求这个式子啊!cosxcos2xcos4xcos8x=?还有一个题 已知2sinx+3siny=a ,2cosx - 3cosy=b,求cos(x+y)? - 作业帮
严光18588658759…… [答案] (sinx*cosxcos2xcos4xcos8x)/sinx sin2xcos2xcos4xcos8x/2sinx sin4xcos4xcos8x/4sinx sin8x*cos8x/8sinx sin16x/16sinx
@沙馥800：计算cosxcos2xcos3x….cosnx
严光18588658759…… cosx+cos2x+cos3x+......+cosnx =1/(2sin(x/2))*[2cosxsin(x/2)+2cos2xsin(x/2)+......+2cosnxsin(x/2)] 括号中的数列的和等于“ [sin(3x/2)-sin(x/2)]+[sin(5x/2)-sin(3x/2)]+[sin(7x/2)-sin(5x/2)]+...... +{sin[(2n+1)x/2]-sin[(2n-1)x/2]} =sin[(2n+1)x/2]-sin(x/2) =2cos(n+1)xsin(nx/2) 所以,原式=cos(n+1)xsinnx/sin(x/2).
@沙馥800：∫cosxcos2xcos3xdx -
严光18588658759…… cosxcos2xcos3x = cos2x * cosxcos3x = cos2x * (1/2)[cos(x + 3x) + cos(x - 3x)] = cos2x * (1/2)[cos4x + cos2x] = (1/2)cos2xcos4x + (1/2)cos²2x = (1/2)(1/2)[cos(2x + 4x) + cos(2x - 4x)] + (1/2)(1/2)(1 + cos4x) = (1/4)cos6x + (1/4)cos2x + 1/4 + (1/4)cos4...
@沙馥800：1.(cosx)^3={(cos2x+1)*(cosx)}/2 是怎么化简得来的?要步骤! -
严光18588658759…… 1,用了个二倍角公式cos2x=2(cosx)^2-1 (cosx)^2=(cos2x+1)/2 2,应该是cos2xcosx=(cos3x+cosx)/2 这个采用了积化和差公式cos3x=cos(2x+x)=cos2xcosx- cosx=cos(2x-x)=cos2xcosx+sin2xsinx 两式相加得到2cos2xcosx=cos3x+cosx
@沙馥800：cosx+cos2x+cos3x+....+cosnx= -
严光18588658759…… 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 ...
@沙馥800：cosxcos2x=1/2(cosx+cos3x) 这个没看懂 求大神解释.. -
严光18588658759…… 积化和差公式现在的教材中已经删除了,可以用下面的替代: 思路分析: 找到角 x与3x 的平均值;2x 让平均值 2x 出场,以平均值为主线;更改原来的角的样式过渡到左边的角 x,及2x, 思路启蒙于等差数列; cosx+cos3x=cos(2x-x)+cos(2x+x) =[cos2xcosx+sin2xsinx]+[cos2xcosx-sin2xsinx] =2cos2xcosx 两边同除以 2得: cosxcos2x=(1/2)(cosx+cos3x)
@沙馥800：∫cosxcos2xcos3xdx - 作业帮
严光18588658759…… [答案] cosxcos2xcos3x = cos2x * cosxcos3x = cos2x * (1/2)[cos(x + 3x) + cos(x - 3x)] = cos2x * (1/2)[cos4x + cos2x] = (1/2)cos2xcos4x + (1/2)cos²2x = (1/2)(1/2)[cos(2x + 4x) + cos(2x - 4x)] + (1/2)(1/2)(1 + cos4x) = (1/4)cos6x + (1/4)cos2x + 1/4 + (1/4)cos4x ∫ ...
@沙馥800：当X趋向0时,1 - cos(X)cos(2X)cos(3X)对于X的无穷小的阶等于?即问与X的几次方是同阶无穷小. - 作业帮
严光18588658759…… [答案] A= [1-cosxcos2xcos3x]/x^2用洛必达法则分子分母求导 A1= [sinx cos2xcos3x + 2cosx sin2xcos3x+3 cosxcos2xsin3x ]/(2x)分子分母求导 A2= [cosx cos2xcos3x -2sinx sin2xcos3x -3sinx cos2xsin3x - 2sinx sin2xcos3...
