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cosx根号cos2xcos3x
高数问题!!
答:
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
cosxcos2x
用积化和差的方法化简
答:
cos(A+B)=cosA.cosB - sinA.sinB (1)cos(A-B)=cosA.cosB + sinA.sinB (2)(1)+(2)cosA.cosB =(1/2) [ cos(A-B)+cos(A+B) ]A=x , B=2x
cosx
.
cos2x
=(1/2) [ cosx+cos(
3x
) ]
当X趋向于0时,1-
cosx
,
cos2x
,
cos3x
与ax^n为等价无穷小,求n,a
答:
1-
cosxcos2xcos3x
=1-cosx(1-2sin^2x)(cosxcos2x-sinxsin2x)=1-cosx(1-2sin^2x)[cosx(1-2sin^2x)-2sin^
2xcosx
]=1-cos^2x(1-2sin^2x)(1-4sin^2x)=1-(1-sin^2x)(1-2sin^2x)(1-4sin^2x)=8sin^6x-14sin^4x+7sin^2x,由于sinx与x为等阶无穷小,而sin^6x和sin^4x相对于...
当X趋向于0时,1-
cosx
,
cos2x
,
cos3x
与ax^n为等价无穷小,求n,a
答:
1-
cosxcos2xcos3x
=1-cosx(1-2sin^2x)(cosxcos2x-sinxsin2x)=1-cosx(1-2sin^2x)[cosx(1-2sin^2x)-2sin^
2xcosx
]=1-cos^2x(1-2sin^2x)(1-4sin^2x)=1-(1-sin^2x)(1-2sin^2x)(1-4sin^2x)=8sin^6x-14sin^4x+7sin^2x,由于sinx与x为等阶无穷小,而sin^6x和sin^4x相对于...
如何把
cosx
+
cos2x
化简为2(
cos3x
/2)(cos1x/2)?(其中x=3x/2-1x/2)
答:
和差化积公式
cosx
+
cos2x
=cos(3x/2-x/2)+cos(3x/2+x/2)=
cos3x
/2cosx/2+sin3x/2sinx/2+cos3x/2cosx/2-sin3x/2sinx/2 =2cos3x/2cosx/2 望采纳
已知函数f( x)=
cos2x
+
cos3x
,则f(1)=?
答:
=1-(1/4)(cos6x+cos2x)-(1/4)(1+cos4x)= 3/4 - (1/4)(cos6x+cos4x+cos2x)lim<x→0>(1-
cosxcos2xcos3x
)/(ax^n)= lim<x→0>[3/4 - (1/4)(cos6x+cos4x+cos2x)]/(ax^n) (0/0)= lim<x→0>(1/2)(3sin6x+2sin4x+sin2x)]/[anx^(n-1)] (0/0)=...
cos2xcos3x
的极限是多少?
答:
cos2xcos3x
=(1/2)(2cos2xcos3x)=(1/2)(2cos2xcos3x+sin2xsin3x-sin2xsin3x)=(1/2)[(cos2xcos3x+sin2xsin3x)+(cos2xcos3x-sin2xsin3x)]=(1/2)[cos(3x+2x)+cos(3x-2x)]=1/2(cos5x+
cosx
)看在手打了这些的份上,请采纳~
y’’=
cos2xcos3x
求通解
答:
∵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)...
求和Sn=
cosx
+
cos2x
+
cos3x
+……+cosnx
答:
+2sin(x/2)cosnx =sin(3x/2)-sin(x/2)+sin(5x/2)-sin(3x/2)+sin(7x/2)-sin(5x/2)+……+sin(x/2+nx)-sin(nx-x/2)=sin(x/2+nx)-sin(x/2)所以
cosx
+
cos2x
+
cos3x
+……+cosnx =[sin(x/2+nx)-sin(x/2)]/[2sin(x/2)]=sin(x/2+nx)/[2sin(x/2)]-1/2 ...
cosxcos2x
=1/2(cosx+
cos3x
) 这个没看懂 求大神解释..
答:
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]=2cos2x...
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