证明sin(x^2 y^2)一致连续

定义g(x)如下g(0)=1g(x)=f(x)=sinx/x(0

说的应该是“一致连续”.　　证明　　（1）注意到　　|[(sinx1)^2]-[(sinx2)^2]|=|sinx1-sinx2|*|sinx1+sinx2|　　0,取δ=ε/2>0,对任意的x1,x

[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-

sin(x+y)-sinx=sinxcosy+cosysinx-sinx2cos((2x+y)\2)sin(y\2)=2cos(x+y/2)sin(y/2)=2[cosxcos(y/2)-sinxsi

【证明】首先必须了解和差化积公式sinα+sinβ=2sin[(α+β)/2]·cos[(α-β)/2]（1）sinα-sinβ=2cos[(α+β)/2]·sin[(α-β)/2]（2）cosα+c

sin(x+y)sin(x-y)=-1/2(cos(x+y+x-y)—cos(x+y-x+y))=-1/2(cos2x—cos2y)=-1/2(1-2（sinx）^2-1+2(siny）^2)=（si

这个是和差化积公式如没学过,可以这样sin(x+y)-sinx=sin[(x+1/2y)+1/2y]-sin[(x+1/2y)-1/2y]=sin(x+1/2y)cos(1/2y)+cos(x+1/2

sin(x+y)sin(x-y)=-1/2(cos(x+y+x-y)—cos(x+y-x+y))=-1/2(cos2x—cos2y)=-1/2(1-2（sinx）^2-1+2(siny）^2)=（si

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

奇函数.f(x)＝lg[sinx+√(1+sin^2x)]因为[－sinx+√(1+sin^2x)]×[sinx+√(1+sin^2x)]＝1,所以,－sinx+√(1+sin^2x)＝1/[sinx

周期为T的函数满足：f(x)=f(x+T)1、如果y=sin[x^2]是周期函数,设最小正周期为T则：sin[x^2]=sin[(x+T)^2]x^2=(x+T)^2+2kπ化简得：2Tx+T^2+2

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

左边=(sinxcosy+cosxsiny)(sinxcosy-cosxsiny)=sin²xcos²y-cos²xsin²y=sin²x(1-sin

f(x)=sin(x^2)＝（1-cos2x)/2因为cos2x在R上可以取得唯一对应值,所以（1-cos2x)/2在r上可以是可以取得唯一对应值的,持续的.

当x≥0时,sin|x/2|=sin(x/2),而sin(x/2)的最小正周期为4π；当x＜0时,sin|x/2|=sin(-x/2)=-sin(x/2),-sin(x/2)的最小正周期也是4π；当-

sinX+sin(X+Y)+sin(X+2Y)/cosX+cos(X+Y)+cos(X+2Y)=sinX+sin(X+2Y)+sin(X+Y)/cosX+cos(X+2Y)+cos(X+Y)=2sin

首先,sin(x)在[0,1]连续故一致连续.即对任意ε>0,存在δ>0,使x,y∈[0,1]满足|x-y|总有|sin(x)-sin(y)|于是对任意a,b∈[1,+∞)满足|a-b|由1/a,1/

sinx+siny+sinz-sin(x+y+z)=4sin[(x+y)/2]sin[(x+z)/2]sin[(y+z)/2]sinx+siny+sinz-sin(x+y+z)=2sin[(x+y)/

题目应该是“证明cosx(cosx-cosy)+sinx(sinx-siny)=2sin²(x-y)/2”Pr：左边展开得cos²x-cosxcosy+sin²x-sin

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