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cosxcos2xcos3x的值是多少
∫
cosxcos2xcos3x
dx
答:
= (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 ∫
cosxcos2xcos3x
dx = (1/4)∫ dx + (1/4)∫ cos2x dx + (1/4)∫ cos4x ...
∫
cosx cos2xcos3x
dx
答:
【1】积化和差:2cosxcos3x=cos4x+cos2x4
cosxcos2xcos3x
=2cos2x[cos4x+cos2x]=cos6x+cos2x+cos4x+1【2】可设原式=y4y=∫[cos6x+cos4x+cos2x+1]dx=(1/6)∫cos6xd(6x)+(1/4)∫cos4xd(4x)+(1/2)∫cos2xd(2x)+∫dx=[(sin...
cos3x
和
cosx的
关系
答:
cos3x
=cos(2x+x)=
cos2xcosx
-sin2xsinx=(2
cos_x
-1)cosx-2(1-cos_x)cosx=4cos_x-3cosx三倍角公式:sin(3α) = 3sinα-4sin3α = 4sinα·sin(60°+α)sin(60°-α)cos(3α) = 4cos3α-3cosα = 4cosα·cos(60°+α)cos(60°-α)tan(3α) = (3tanα-...
为什么
cos3xcos2x
=1/2(cos5x+
cosx
)
答:
cos2xcos3x
=cos3xcos2x =(1/2){cos(3x+2x)+cos(3x-2x)} =1/2(cos5x+
cosx
)
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...
当x→0时,1-
cosxcos2xcos3x
与a
x的
n次幂为等阶无穷小,求n与a
的值
哪位...
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...
当x→0时,1-
cosxcos2xcos3x
与axn是等价无穷小,求常数a,n.用泰勒展开co...
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...
已知函数f(x)=1/2(
cosx
),求实数a的取值范围。
答:
1-
cosxcos2xcos3x
= 1-(1/2)cos2x(cos4x+cos2x)= 1-(1/2)cos2xcos4x-(1/2)(cos2x)^2 =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)]/...
cosxcos2xcos3x是
怎样转化成(cos6x+cos4x+cos2x+1)/4 的,求过程
答:
摆渡一下积化和差公式。和差化积公式。或者用我这种整体法等价无穷小逆向思维双向思维先写别问。,对数是logarithm的log或者LNX,Lg绝非ig,并非inx,不是logic缩写,反民科吧,恒等式π^a=exp(Ln(π^a))=e^(alnπ)。对不起打扰了唉。abs绝对值,sqrt开根号。举报数字帝国GG泛滥但是是一个计算器...
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