y=(2cosx 1) (2cosx-1)值域
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/03 11:10:40
设sinθ-cosθ=√2sin(x-π/4)=t则:t属于[-1,√2]sinxcosx=(1-t^2)/2y=-t^2+t+1=-(t-1/2)^2+5/4最大值是:5/4(此时t=1/2)最小值
sin(x+y)sin(x-y)=[sinxcosy+sinycosx][sinxcosy-cosxsiny]=(sinxcosy)^2-(cosxsiny)^2=(1-cos^2y)cos^2y-c
y=cos^2x-2cos^2(x/2)=cos^2x-cosx-1=(cosx-1/2)^2-5/4一个单调增区间[-π/3,0]再问:答案是(π/3,π)再答:(0,π/3]单减[π/3,π/2]
由y=sinx+cosx1+sinx,得y+ysinx=sinx+cosx,即(y-1)sinx-cosx=-y,∴(y−1)2+1sin(x+φ)=-y,则sin(x+φ)=−y(y−1)2+1,∵
设t=cos(x/2),-11/2时,单调递增.cos(x/2)>1/2,-PI/3-2PI/3因此,函数y=cos^x-2cos^(x/2)的一个单调增区间为,-2PI/3
COS(X+Y)COS(X-Y)=(COSX*COSY-SINX*SINY)(COSX*COSY+SINX*SINY)=(COSX*COSY)^2-(SINX*SINY)^2=COS^2X(1-SIN
5/9cos(x-y)=cosx*cosy+sinx*sinysin(x)-sin(y)=-(2/3),两边平方得到sin^2x-2sinxsiny+sin^2y=4/9cos(x)-cos(y)=(
再问:晕死,,我连这种题目都不会化再问:我想死了😱😱😖再答:会了么现在再问:懂了,再问:我发现我很笨再答:没有啦,我刚学的那会也不会再答:亲给好评
∵y=1−2cosx1+2cosx,∴cosx=1−y2+2y,∵-1≤cosx≤1,∴|cosx|=|1−y2+2y|≤1,即(1-y)2≤(2+2y)2,解得:y≤-3或y≥-13,∴函数y=1−
y'=(cos²x)'-(sin3^x)'=2cosx·(cosx)'-cos3^x·(3^x)'=2cosx·(-sinx)-cos3^x·(3^x·ln3)=-sin2x-ln3·cos
函数y=cosπ/2x×cosπ/2(x-1)的最小正周期如果是:函数y=cosπ/2x×cos[(π/2)(x-1)]的最小正周期则有如下:y=cosπ/2x×cosπ/2(x-1)=cosπx/2
y=[cosx-1-1]/(cosx-1)=1-1/(cosx-1)=1-1/(1-2sin^2(x/2)-1)=1+1/(2*sin^2(x/2))故其周期是T=2π
-2k=cos2x-cos2y=[2(cosx)^2-1]-[2(cosy)^2-1]=2[(cosx)^2-(cosy)^2]cos^2x-cos^2y=-k
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
y=cosx^2y'=2cosx(COSX)'=-2SINXCOSXy=cos2xy'=-SIN2X(2X)'=-2SIN2X
y=(sinx+cos)^2+2cos^2x=1+2sinxcosx+cos2x-1=sin2x+cos2x=√2sin(2x+π/4)
由于函数y=cosx1−sinx=cos2x2−sin2x2cos2x2+sin2x2−2sinx2cosx2=1−tan2x21+tan2x2−2tanx2=(1+tanx2)(1−tanx2)(1
你可以把分母sinx-siny用和差化积化成2sin((x-y)/2)cos((x+y)/2)这样答案就很显然了
y=sin²x+2sinxcosx+3cos²xy=(sin²x+cos²x)+2sinxcosx+(2cos²x-1)+1=1+sin2x+cos2
y'=2cos(sin2x)×[cos(sin2x)]'=2cos(sin2x)×[-sin(sin2x)]×(sin2x)'=-sin(2sin2x)×2cos2x=-2cos2xsin(2sin2