比较sin^3x cos^4x x^2sin^3x - cos^4x

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比较sin^3x cos^4x x^2sin^3x - cos^4x
设函数f(x)=根号3cos²ωx+sinωxcosωx(ω,a∈R),

f(x)=根号3cos²ωx+sinωxcosωx=根号3/2(cos²2ωx)+1/2(sin2ωx)=sin(2ωx+π/3);y轴右侧的第一个最高点的横坐标为π/6,π/3ω

已知函数f(x)=4sinωxcos(ωx+π3)+3(ω>0)的最小正周期为π.

(1)∵f(x)=4sinωx(cosωxcosπ3−sinωxsinπ3)+3,------(1分)=2sinωxcosωx−23sin2ωx+3=sin2ωx+3cos2ωx---(3分)=2si

已知函数f(x)=sin^2ωx+√3cosωxcos(π/2-ωx)(ω>0)

f(x)=sin^2ωx+√3cosωxcos(π/2-ωx)(ω>0)=(1-cos2ωx)/2+(√3/2)sin2ωx=sin(2ωx-π/6)+1/2∵函数y=f(x)的图像相邻两条对称轴之间

sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x) =sin^4x-sin^2xcos

那个前半括号里面相加等于一

化简cos^4x+sin^2xcos^2x+sin^2x

cos^4x+sin^2xcos^2x+sin^2x=cos^4x+(1-cos²x)cos²x+sin²x=cos^4x+cos²x-cos^4x+sin&#

求函数f(x)=(sin^4x+cos^4x+sin^2xcos^2x)/2

sin^4x+cos^4x+sin^2x*cos^2x=sin^4x+cos^4x+2sin^2x*cos^2x-sin^2x*cos^2x=(sin^2x+cos^2x)^2-sin^2x*cos^

数学题目已知函数f(x)=+2sinωxcosωx+2f(a)=2/3求sin(5/6π-4a)

f(x)=sin2ωx+√3cos2ωx=2sin(2ωx+π/3),两对称轴之间的最小值为π/2即半个周期,则周期为π=2π/2ω,所以w=1,所以f(x)=2sin(2x+π/3),f(α)=2s

求证 cos*xcos*y + sin*xsin*y + sin*xcos*y + xin*ycos*x = 1

合并同类项么,很简单的只要你愿意去做左边=cos*x(cos*y+sin*y)+sin*x(cos*y+sin*y)=cos*x+sin*x=1=右边

问高数求导 ∫sin^3xcos^2xdx

∫sin^3xcos^2xdx=-∫sin^2xcos^2xdcosx=-∫(1-cos^2x)*cos^2xdcosx=-∫(cos^2x-cos^4x)dcosx=(1/5)*cos^5x-(1/

已知Y=sin^4*3xcos^3* 4x求Y的导数

y=sin⁴3xcos³4xdy/dx=cos³4x*d(sin⁴3x)/dx+sin⁴3x*d(cos³4x)/dx=cos

化简sin^2(a+π)Xcos(π+a)Xcot(-a-2π)/tan(π+a)Xcos^3(-a-π)

sin^2(a+π)Xcos(π+a)Xcot(-a-2π)/tan(π+a)Xcos^3(-a-π)=sin^2a(-cosa)(-cota)/tana(-cos^3a)=-sinacos^2a/s

3道不定积分数学题求下列不定积分(1)cos2t/cost –sint dt(2)cos2x/sin^xcos^x dx

看:(对不起,第一条的变数全部都是t,刚才做的时候不小心把t打错作x了)

求函数y=2sin xcos x+2sin x+2cos x+4的值域

t=sinx+cosx=√2sin(x+π/4)-√2=再问:上面那个颠倒的V是什么再答:那是根号呀,√2表示根号2.再问:sin^2x这个颠倒的^也是根号?再答:这个是次方符号呀,sin^2x表示的

求sin^4x+cos^4x+4sin^2xcos^2x-1的最小正周期及值域.

y=(sin^2x+cos^2x)^2+2sin^2xcos^2x-1=1+2sin^2xcos^2x-1=2sin^2xcos^2x=sin^2(2x)/2=(1-cos4x)/4周期显然是pi/2

求定积分∫上限π/2,下限0 4sin^2xcos^2xdx,

这题方法有很多,你可以把cos^2x换成1-sin^2x4sin^2xcos^2x=4(sin^2x-sin^4x)sin^2x和sin^4x积分是有公式的.但是一般人估计也记不得,所以方法二:为了方

x趋于0时,lim(x^2-sin^2 xcos^2 x)/(x^2sin^2 x)怎么转换成(x^2-(1/4)sin

2sinxcosx=sin2x那么sin^2xcos^2x=sin^22x/4另外sinx等价于x,所以sin^2x等价于x^2,也即x^2sin^2x变成了x^4不知您是否明白,若有不明还可问(⊙o

求证 sinˇ4X+sin²Xcos²X+cos²X = 1

证明:因为左边=sin²X(sin²X+cos²X)+cos²X=sin²X+cos²X=1=右边,所以:(sinX)^4+sin²

(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x

sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x

∫sin²xcos³x dx

∫sin^2xcos^3xdx=∫sin^2x(1-sin^2x)dsinx=∫sin^2x-sin^4xdx=(1/3)sin^3x-(1/5)sin^5x+C不是让你求助我吗.再问:∫sin^2x