y=cos(x 兀 6) x∈[0,兀 2]值域

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y=cos(x 兀 6) x∈[0,兀 2]值域
已知cos(x)=1/7 cos(x+y)=-11/14 且x,y属于(0,90) 求cos(y)

因为sin^x+cos^x=1x,y属于(0,90),所以sinx>0sinx=4*根号3/7cos^(x+y)+sin^(x+y)=1sin(x+y)>0所以sin(x+y)=5*根号3/14因为c

已知TanX,TanY是方程X^-3X-3=0的俩根,求sin^(x+y)-3sin(x+y)cos(x+y)-3cos

tanx+tany=3(tanx)(tany)=-3tan(x+y)=(tanx+tany)/(1-tanxtany)=3/4[sin(x+y)]^2+[cos(x+y)]^2=1[sin(x+y)]

已知函数 y=sin(2x-(π/3))+cos(2x-(π/6))+2cos²x-1,x∈R

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

证明COS(X+Y)COS(X-Y)=COS^2X-SIN^2Y

COS(X+Y)COS(X-Y)=(COSX*COSY-SINX*SINY)(COSX*COSY+SINX*SINY)=(COSX*COSY)^2-(SINX*SINY)^2=COS^2X(1-SIN

求函数y=sin^4x+cos^4x,x(0,π/6)的最小值

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

Matlab编程问题 cos(x*y)*cos(x*(1-y))-0.5x*sin(x*y)*sin(x*(1-y))=

symsxyeq=cos(x*y)*cos(x*(1-y))-0.5*x*sin(x*y)*sin(x*(1-y))-1;ezplot(eq)

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

已知函数y=4 cos²x+4倍根号3 sin x cos x-2,x∈R.

y=2(2cos²x-1)+2倍根号三sin2xy=2cos2x+2倍根号三sin2xy=4(1/2倍cos2x+根号三/2倍sin2x)y=4sin(π/6+2x)三角函数解析式有了想要什

已知函数y=cos^4x-2sinxcosx-sin^4x,x∈[0,∏/2]

1y=cos2x-sin2x=2sin(2x-∏/4)最大值是222k∏+∏/2∠(2x-∏/4)∠2k∏+3/2∏k∏+3/8∏∠x∠k∏+7/4∏32sin(2x-∏/4)>-1sin(2x-∏/

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

设f(x)=4cos(ωx-π/6)sinωx-cos(2ωx+π),其中ω>0,(1)求函数y=f

这个用cos(α-β)好想可以做出来,最好问老师

函数y=sin(x+兀/2)cos(x+兀/6)的递减区间是

sin(x+π/2)=cosx然后打开得到f(x)=cosx(√3/2cosx-1/2sinx)=√3/2(cosx)^2-1/2cosxsinx再用二倍角公式得到1/4sinx+√3/4cosx+√

y=(cosα)x²-(2sinα)x+6,任意x∈R,都有y>0,求cosα取值范围?

y恒大于0,则抛物线y须开口向上,即cosα>0(i);且∆=4sin²α-24cosαsin²α-6cosα=1-cos²α-6cosαcos²α

绘制x=sin(x),y=cos(x),z=t在[0,6pi]上的三维曲线

symst>>x=sin(t);>>y=cos(t);>>z=t;>>ezplot3(x,y,z,[0,6*pi])用的是matla

函数y=cos(x+π/6),x属于【0,π/2】的值域是

forx=π/2y=cos(π/2+π/6)=cos(2π/3)=-1/2x属于【0,π/2】的值域是[-1/2,1]

cos(x+y)=?sin(x+y)=?

cos(x+y)=cosxcosy-sinxsinysin(x+y)=sinxcosy+cosxsiny

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

) y=cos(x-y)

1.两边求导得:y'=-sin(x-y)(1-y')解得y'=sin(x-y)/[sin(x-y)-1]2.y'=-e^-xy''=e^-xy'"=-e^-x3.y'"=(e^2x)'"(sinx)+