cosxcos2xcos3x化简
@储瑗5346：cosxcos2xcos3x的不定积分 - 作业帮
漆娇13053839919…… [答案] 你好!数学之美团为你解答积化和差公式:cosαcosβ = 1/2 [ cos(α+β) + cos(α - β) ]cosx cos2x cos3x= 1/2 ( cos3x + cosx ) cos3x= 1/2 cos²(3x) + 1/2 cosx cos3x= 1/4 ( 1 + cos6x ) + 1/4 ( cos4x + c...
@储瑗5346：求cosxcos2xcos3x对x的不定积分,及这种很多三角相乘除的方法. - 作业帮
漆娇13053839919…… [答案] 先用积化和差公式化简得1/4(1+cos6x+cos4x+cos2x)再分部积之得1/4x+1/24sin6x+1/16sin4x+1/8sin2x+C
@储瑗5346：∫cosxcos2xcos3xdx - 作业帮
漆娇13053839919…… [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 ∫ ...
@储瑗5346：(1 - cosxcos2xcos3x)/(1 - cosx)当x趋近于0时的极限 -
漆娇13053839919…… 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...
@储瑗5346：cosxcos2xcos3x的导数原题1 - cosxcos2xcos3x/1 - cosx 在x趋近于0的极限 - 作业帮
漆娇13053839919…… [答案] 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
@储瑗5346：求cosxcos2xcos3x对x的不定积分,及这种很多三角相乘除的方法. -
漆娇13053839919…… 先用积化和差公式化简得1/4(1+cos6x+cos4x+cos2x)再分部积之得1/4x+1/24sin6x+1/16sin4x+1/8sin2x+C
@储瑗5346：计算cosxcos2xcos3x….cosnx
漆娇13053839919…… 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).
@储瑗5346：∫cosxcos2xcos3xdx -
漆娇13053839919…… 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...
@储瑗5346：y''=cos2xcos3x求通解 -
漆娇13053839919…… ∵y″=cos2xcos3x=(1/2)(cosx+cos5x), ∴y′=(1/2)∫cosxdx+(1/2)∫cos5xdx=(1/2)sinx+(1/10)sin5x+A, ∴y=(1/2)∫sinxdx+(1/10)∫sin5xdx+A∫dx=-(1/2)cosx-(1/50)cos5x+Ax+B. ∴原微分方程的通解是:y=-(1/2)cosx-(1/50)cos5x+Ax+B,其中A、B为任意实数.
@储瑗5346：当x→0时,1 - cosxcos2xcos3x与axn是等价无穷小,求常数a,n. - 作业帮
漆娇13053839919…… [答案] 当x→0时余弦函数在x=0的带佩亚诺余项的泰勒展开式: cosx= n+1 k=1 (−1)k−1x2k−2 (2k−2)!+o(x2n) 则当x→0时函数在x=0的带佩亚诺余项的二阶泰勒展开式分别为: cosx=1− 1 2x2+o(x2) cos(2x)=1− 1 2(2x)2+o(x2)=1-2x2+o(x2) cos(3x)=1− 1 ...
漆娇13053839919…… [答案] 你好!数学之美团为你解答积化和差公式:cosαcosβ = 1/2 [ cos(α+β) + cos(α - β) ]cosx cos2x cos3x= 1/2 ( cos3x + cosx ) cos3x= 1/2 cos²(3x) + 1/2 cosx cos3x= 1/4 ( 1 + cos6x ) + 1/4 ( cos4x + c...
@储瑗5346：求cosxcos2xcos3x对x的不定积分,及这种很多三角相乘除的方法. - 作业帮
漆娇13053839919…… [答案] 先用积化和差公式化简得1/4(1+cos6x+cos4x+cos2x)再分部积之得1/4x+1/24sin6x+1/16sin4x+1/8sin2x+C
@储瑗5346：∫cosxcos2xcos3xdx - 作业帮
漆娇13053839919…… [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 ∫ ...
@储瑗5346：(1 - cosxcos2xcos3x)/(1 - cosx)当x趋近于0时的极限 -
漆娇13053839919…… 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...
@储瑗5346：cosxcos2xcos3x的导数原题1 - cosxcos2xcos3x/1 - cosx 在x趋近于0的极限 - 作业帮
漆娇13053839919…… [答案] 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
@储瑗5346：求cosxcos2xcos3x对x的不定积分,及这种很多三角相乘除的方法. -
漆娇13053839919…… 先用积化和差公式化简得1/4(1+cos6x+cos4x+cos2x)再分部积之得1/4x+1/24sin6x+1/16sin4x+1/8sin2x+C
@储瑗5346：计算cosxcos2xcos3x….cosnx
漆娇13053839919…… 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).
@储瑗5346：∫cosxcos2xcos3xdx -
漆娇13053839919…… 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...
@储瑗5346：y''=cos2xcos3x求通解 -
漆娇13053839919…… ∵y″=cos2xcos3x=(1/2)(cosx+cos5x), ∴y′=(1/2)∫cosxdx+(1/2)∫cos5xdx=(1/2)sinx+(1/10)sin5x+A, ∴y=(1/2)∫sinxdx+(1/10)∫sin5xdx+A∫dx=-(1/2)cosx-(1/50)cos5x+Ax+B. ∴原微分方程的通解是:y=-(1/2)cosx-(1/50)cos5x+Ax+B,其中A、B为任意实数.
@储瑗5346：当x→0时,1 - cosxcos2xcos3x与axn是等价无穷小,求常数a,n. - 作业帮
漆娇13053839919…… [答案] 当x→0时余弦函数在x=0的带佩亚诺余项的泰勒展开式: cosx= n+1 k=1 (−1)k−1x2k−2 (2k−2)!+o(x2n) 则当x→0时函数在x=0的带佩亚诺余项的二阶泰勒展开式分别为: cosx=1− 1 2x2+o(x2) cos(2x)=1− 1 2(2x)2+o(x2)=1-2x2+o(x2) cos(3x)=1− 1 ...
