求函数y=sin²x 根3Cos X 5 4的最大值和最小值

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)]

y=sin²x+cos8x+2sinxcosx+2cos²x=1+sin2x+(1+cos2x)=sin2x+cos2x+2=√2sin(2x+π/4)+2-1

对x求导1*cosy+x*(-siny)*y'=cos(x+y)*(x+y)'cosy-xsiny*y'=cos(x+y)(1+y')=cos(x+y)+cos(x+y)*y'所以y'=[cosx-c

y=x-sinx/2cosx/2y=x-sin(x/2)cos(x/2)=x-(1/2)sinxdy/dx=1-(1/2)cosxy=x^3+3^xy=x^3+3^x,y'=3x^2+3^xln3

y=sinxcos30+cosxsin30-cosxsin60-sinxcos60=sinx[(根号3-1)/2]+cosx[(1-根号3)/2]=[(根号3-1)/2](sinx-cosx)=[(根

求函数y=sin²x+3sinxcosx+5cos²x的值域y=1-cos²+3sinxcosx+5cos²x=4cos²x+3sinxcosx+1=

整理方程,得y=1+2sinxcosx+2(cosx)^2利用降幂公式和二倍角公式,得y=sin2x+cos2x+2再利用辅助角公式,得y=根号2*sin(2x+π/4)+2所以当2x+π/4属于[2

y=(sin²x+cos²x)+(√3/2)(2sinxcosx)+cos²x=1+(√3/2)sin2x+(1+cos2x)/2=(√3/2)sin2x+(1/2)co

y=1+√3/2sin2x+1/2+1/2cos2x=3/2+sin(2x+π/6)【注：π是派--】∴周期T=2π/2=π值域=[1/2,5/2]

sin3x=sin(x+2x)sin^3x=(sinx)^2*sinx(sinx)^2=(1-cox2x)/2sin3x*sin^3x化简得0.5*(cos2x-2(cos2x)^2+1)*((1-c

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

y=sinx+√3cosx=2sin(x+π/3)最小正周期为2π最大值为2,最小值为-2

派再问：学霸过程呢！再答：再答：刚刚在吃饭，随便瞄了一眼，现在刚回图书馆，不好意思让久等了，不知道对不对，半年不做写些题目了，有些生疏再答：果真漏掉个负号～～不过答案应该没错再问：我会告诉你我没看懂-

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=2sin²x+cosx+3=-2cos²x+cosx+5=-2(cosx-1/4)²+41/8cosx=1/4,max(y)=41/8cosx=-1,min(y)=2

原式=(1-cos2x)/2+(sin2x)/2+2=(sin2x-cos2x)/2+5/2=(sin（2x-45度）)*(根号2)/2+5/2所以是大于（根号2+5）/2,小于（5-根号2）/2

正在做啊再问：恩再答：cos[π/2-(π/3+x)]=cos(π/6-x)=sin(π/3+x)y=sin(x+π/3)cos(π/6-x)=sin(x+π/3)sin(π/3+x)=sin

从周期的定义来看,令f(x)=sin3x+cos5x,f(x+2Pi)=sin(3x+6Pi)+cos(5x+10Pi)=sin3x+cos5x=f(x)所以周期为2Pi,其实只要找到一个T即是sin

y=cos²x-sin²x+2sinxcosx=cos2x+sin2x=√2sin(2x+π/4)所以值域为【-√2,√2】

y=1-sin²x-3sinx=-(sinx+3/2)²+13/4对称轴sinx=-3/2,开口向下而-1