y=cos^2x-3sinx

1,y'=-sin(4-3x)·(4-3x)'=3sin(4-3x)2,y=ln(1+x^2)^3=3ln(1+x^2)y'=3/(1+x^2)·(1+x^2)'=6x/(1+x^2)3,令y=x^s

[sin(2x+y)/sinx]-2cos(x+y)={[sin(x+y)cosx+cos(x+y)sinx]/sinx}-2cos(x+y)={[sin(x+y)cosx+cos(x+y)sinx-

|x|

y=(sinx+cosx)^2+2cos^x=2+sin2x+cos2x=2+√2sin(2x+π/4)ymax=2+√2,ymin=2-√2.2kπ+π/2≤2x+π/4≤2kπ+3π/22kπ+π

解y=sin^2X+2sinX*cosX+3COS^2X=sin^2X+COS^2X+2sinX*cosX+2COS^2X=1+sin2x+2COS^2X=2+sin2x+2COS&^2X-1=2+s

y=sin^2x+2sinx*cosx+3cos^2x=sin2x+cos2x+2=根号2sin(2x+pi/4)+2所以最小值是2－根号2次时2x+pi/4=2kpi-pi/2x=kpi-3pi/8

1y=(1+sinx-(cosx)^2)/(1+sinx)=[sinx+(sinx)^2]/(1+sinx)=sinx(sinx≠-1)f(x)=sinxf(-x)=-sinx=-f(x)奇函数2y=

y=cos^2x-2sinx=1-sin^2x-2sinx令sinx=t则y=-t^2-2t+1配方得y=-(t+1)^2+2∵-1＜=t＜=1故函数Ymax=2Ymin=-2综上,此函数的值域为[-

y=1-2(sinx)^2+sinx=-2x^2+x+1(设sinx=x,x属于[-1,1]画图即可最小值f(-1)=-2最大值f(1/4)=9/8值域为[-2,9/8]

Sin[x]^3+Sin[x]Cos[x]^2=Sin[x](Sin[x]^2+Cos[x]^2)=Sin[x]周期2π.

y=cos^2x+sinx=1-2(sinx)^2+sinx=-2(sinx-1/4)^2+9/8因为|x|

y=sinx+cos^2xy=sinx+1-sin^2(x)y=-(sinx-1/2)^2+5/4因为sinx∈[-1,1]所以y∈[-1,5/4]

令y’=-2cosxsinx-3cosx=-cosx(2sinx+3)=0X1=2kπ-π/2,x2=2kπ+π/2y”=-2cos2x+3sinx==>y”(x1)0∴函数y在x1处取极大值3,在x

y=cosx-√3sinx=2(1/2cosx-√3/2sinx)=2(cosxcosπ/3-sinxsinπ/3)=2cos(x+π/3)再问：那也可以化成2sin(x-π/6）吗再答：y=cosx

求值域的吧1；cos²x+sinx-2=1-sin²x+sinx-2=--sin²x+sinx+3=-【sinx-（1/2）】^2+13/4=x【-π/6,2π/3】si

把5siny=sin(2x+y)变为5sin[(x+y)-x]=sin[（x+y)+x],把其中的（x+y）,看成一个整体,上式即变为4sin(x+y)cosx=6cos(x+y)sinx,再把式子的

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

把cos^2x=1-sin^2xy=cos^2x+sinx+1=1-sin^2x+sinx+1=-sin^2x+sinx+2=-(sinx-1/2)^2+9/4最大值为9/4最小值为2

y=(2sinx-1)/(sinx+3)=（2sinx+6-7）/(2sinx+6)=1-7/(2sinx+6)whensinx=1ymax=1/8whensinx=-1ymin=-3/4y的值域是（

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