x=ο,y=sinx-2cosx有最大值

y'=(sinx)'+(cos^5x)'=cosx-5cos^4x*sinx符合函数的求导,其实就是对式子一层一层的求导.对于cos^5x,可以看做a=cosx,b=a^5.再一层层求导就可以了

[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-

设A=(X+Y)/2,B=(X-Y)/2X=A+B,Y=A-BSINX=SIN(A+B)=SINACOSB+COSASINBSINY=SIN(A-B)=SINACOSB-COSASINBSINX+SI

y=(2sinx-cos^2x)/(1+sinx)=[2sinx-1+(sinx)^2]/(1+sinx)=[(1+sinx)^2-2]/(1+sinx)=(1+sinx)-2/(1+sinx)令：1

|x|

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=(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

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]

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]

sinx+siny=sin[(x+y)/2+(x-y)/2]+sin[(x+y)/2-(x-y)/2]=sin(x+y)/2*cos(x-y)/2+cos(x+y)/2*sin(x-y)/2+sin(

|2sin[(x-y)/2]cos[(x+y)/2]|=I2[sin(x/2)cos(y/2)-sin(y/2)cos(x/2)][cos(x/2)cos(y/2)-sin(x/2)(sin(y/2)

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

y=(1+sinx)/cosxy'=[(1+sinx)/cosx]'=[cosx*cosx-(1+sinx)*(-sinx)]/cos^2x=（sinx+1）/cos^2x你算的是对的啊!恭喜恭喜!

y=(sinx+cos)^2+2cos^2x=1+2sinxcosx+cos2x-1=sin2x+cos2x=√2sin（2x+π/4)

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

x+y=x+y/2+y/2x=x+y/2-y/2sin(x+y)=sin(x+y/2+y/2)=sin(x+y/2)cosy/2+cos(x+y/2)siny/2sinx=sin(x+y/2)cosy

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（