求y=4cos(2x+π 3)-3的最值

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求函数y=2cos(x+π4)cos(x−π4)+3sin2x

y=2cos(x+π4)cos(x−π4)+3sin2x=2(12cos2x−12sin2x)+3sin2x=cos2x+3sin2x=2sin(2x+π6)∴函数y=2cos(x+π4)cos(x−

求函数y=log1/2[cos(x/3+π/4)]的单调区间

这是一个复合函数,对于复合函数而言,内外相同则为增,不同则为减在考虑单调区间时,还要注意定义域外层是log1/2a=1/2,是减函数内层减区间2kπ

求函数y=log1/2cos(x/3+π/4)的递减区间

y=log1/2cos(x/3+π/4)t=cos(x/3+π/4)y=log1/2(t)在定义域内是减函数要使y=log1/2cos(x/3+π/4)是减函数t=cos(x/3+π/4)必须是增函数

求函数y=cosx+cos(x-π3

∵y=cosx+cos(x-π3)=cosx+cosxcosπ3+sinxsinπ3=32cosx+32sinx=3(cosπ6cosx+sinπ6sinx)=3cos(x-π6),∵-1≤cos(x

求y=sin2x+2倍根号2cos(π/4+x)+3的最小值

/>y=sin2x+2√2cos(π/4+x)+3=cos(2x-π/2)+2√2cos(π/4+x)+3=1-2sin²(x-π/4)-2√2sin(x-π/4)+3=4-2[sin

1.求函数y=log1/2cos(x/3+π/4)的单调区间 2.求函数y=sin²x+cosx-4(x∈R)

1.求函数y=log1/2cos(x/3+π/4)的单调区间是以1/2为底吗?如果是由于0

求下列函数导数y=cos(π/3-x)y=e^3xy=In(3-x)y=cos^3(1-2x)

y=cos(π/3-x)y'=-sin(π/3-x)*(-1)=sin(π/3-x)y=e^3xy'=e^(3x)*3=3e^(3x)y=In(3-x)y'=1/(3-x)*(-1)=1/(x-3)y

y =(cos^2) x - sin (3^x),求y'

y'=(cos²x)'-(sin3^x)'=2cosx·(cosx)'-cos3^x·(3^x)'=2cosx·(-sinx)-cos3^x·(3^x·ln3)=-sin2x-ln3·cos

sin(x+y)sin(x-y)=k,求cos^2x-cos^2y

-2k=cos2x-cos2y=[2(cosx)^2-1]-[2(cosy)^2-1]=2[(cosx)^2-(cosy)^2]cos^2x-cos^2y=-k

Sin x-sin y=2/3 cos x-cos y=1/2 求cos(x-y)

Sinx-siny=2/3cosx-cosy=1/2分别平方得(Sinx-siny)^2=(2/3)^2(cosx-cosy)^2=(1/2)^2展开相加得-2cos(x-y)+2=4/9+1/4-2

求函数y=3cos(2x+π/4)的对称轴方程

解y=3cosx的对称轴方程是x=kπ.k属于Z即函数y=3cos(2x+π/4)的对称轴方程为2x+π/4=kπ,k属于Z即为x=kπ/2-π/8,k属于Z

求函数的导数y=cos(4-3x)

y'=-sin(4-3X)*(-3)=3sin(4-3X)

求y=sin(2x+π/3)+cos(2x-π/6)的单调区间

y=sin(2x+π/3)+cos(2x-π/6)=(1/2)sin2x+(√3/2)cos2x+(√3/2)cos2x+(1/2)sin2x=sin2x+√3cos2x=2sin(2x+π/3)2k

1.y=cos^4x+sin^4x 求周期 2.y=(sin2x+sin(2x+π/3))/( cos2x+cos(2x

1、y=(cos^2x+sin^2x)^2-2cos^2xsin^2x=1-1/2(sin2x)^2=1-1/4(1-cos4x)=3/4+1/4cos4x周期T=2pi/4=pi/22、y=(根3/

求函数最大值最小值及对应x的集合 y=cos(-x/3+π/4)

最大值为1,当-x/3+π/4=2kπ时取得;最小值为-1,当-x/3+π/4=2kπ+π时取得

求下列函数的周期:(1)y=2cos(2x+π/4)(2)y=cos(3x/5)(3)y=2cos(π/4-x/3)

T=2π/w=2π/2=πT=2π/w=2π/(3/5)=10π/3T=2π/w=2π/(1/3)=6π从以上例子可知,周期与初相角无关,周期只与w有关,而计算周期时,w应等于x系数的绝对值

已知y=cos^4(2x+π/3),求dy/dx

y'=4*cos^3(2x+π/3)*[cos(2x+π/3)]'*(2x+π/3)'=-8*cos^3(2x+π/3)*sin(2x+π/3)

求值域y=2sinx+cos^2x,x∈[π/6,2π/3)

y=2sinx+cos^2x=2sinx+1-sin²x=-(sinx-1)²+2已知x∈[π/6,2π/3),那么:sinx∈[1/2,1]所以当sinx=1即x=π/2时,函数

求函数y=4cos(x/3),0

y=4cos(x/3)得出:x=3arccos(y/4)∴反函数为:y=3arccos(x/4)反函数的定义域就是原函数的值域,0