(siny² xcoty)y=1
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/29 21:33:43
有cos(x+y)=cosxcosy-sinxsiny又将题目所给式子左右平方,在两个式子左右对应相加为:cos^2x+cos^2y+2cosxcsoy+sin^2x+sin^2y-2sinxsiny
两边对x求两次导数:1-y'+1/2cosyy'=0;==>y'=1/(1-cosy/2)0-y''+1/2(y'(-siny)+cosyy'')=0==>y''=y'siny/(cosy-2)再将y
有cos(x+y)=cosxcosy-sinxsiny又将题目所给式子左右平方,在两个式子左右对应相加为:cos^2x+cos^2y+2cosxcsoy+sin^2x+sin^2y-2sinxsiny
cos(x-y)=cosxcosy+sinxsiny=[1-(sinxsiny)^2]^(1/2)+sinxsiny=0+1=1
cos^2y=1-sin^2(y)sinx=1/3-siny原式=1/3-siny-(1-sin^2y)令siny=t∈[-1,1]则f(t)=t^2-t-2/3,t∈[-1,1]二元一次方程,定义域
sinx+siny=1/3sinx=1/3-siny(siny)^2+(cosy)^2=1(cosy)^2=1-(siny)^2u=sinx-cos^2yu=(1/3-siny)-[1-(siny)^
sinx+siny=1/3sinx=1/3-siny-1≤sinx≤1-1≤1/3-siny≤1-2/3≤siny≤4/3又-1≤siny≤1,因此-2/3≤siny≤1sinx-cos²y
【1】sinx+siny=1/4左右平方,【2】cosx+cosy=1/3左右各平方,【1】【2】相加可求出cos{x-y}那么sin[x-y]2+cos[x-y]2=1可求出sin[x-y]的值上下
∫e^ysinydy=-∫e^yd(cosy)=-[e^y*cosy-∫cosyd(e^y)]=∫cosy*e^ydy-e^ycosy=∫e^yd(siny)-e^ycosy=e^ysiny-∫sin
∫(0→1)dx∫(x→1)(siny)/ydy,交换积分次序=∫(0→1)(siny)/ydy∫(0→y)dx=∫(0→1)(siny)/y·ydy=∫(0→1)sinydy=-cosy:[0→1]
∫(y^2→y)siny/ydx=[siny/yx]|(y^2→y)=(y-y^2)siny/y这里是把siny/y看成常数来积分再问:为什么可以看做常数?再答:因为这里x,y是两个自变量,互不相关,
解arcsiny=x中y是自变量,x是因变量∴(arcsiny)'=x'=1/√(1-y^2)≠1例如y=sinx,(sinx)'=y'≠1
答案是1-sin(1)再问:嗯,是的,请问过程?再答:看网页,有图片的他那个是√x到x你那个是x到√x上下限交换就可以了再问:嗯,好的,谢谢啦。∫(-1→1)(x+1)根号下(1-x^2)dx=?请问
原式=cosacosy+sinasiny=cosa(cosa+cosb)+sina(sina+sinb)=1+cosacosb+sinasinbsin^2(y)+cos^2(y)=1+1+2sinas
y=siny+(sinx)^2-1.(sinx^2+cosx^2=1)=1/3-sinx+(sinx)^2-1=(sinx)^2-sinx-2/3=(sinx-1/2)^2-2/3-1/4=(sinx
原条件等价于cos(x-y)=1/3根据倍角公式cos2(x-y)=2cos^2(x-y)-1=2/9-1=-7/9
sinx+siny=2sin(x/2+y/2)·cos(x/2-y/2)=-1/3----①cosx+cosy=2cos(x/2+y/2)·cos(x/2-y/2)=1/2----②①/②=tan(x
首先说明arc(sinx)与1/sinx是没有联系的,跟不会是相同的y=arc(sinx)实际上就是x=siny,比如pi/6=arcsin1/21/2=sinpi/6y=arc(sinx)可以变形为
cosx-cosy=1/4cos²x-2cosxcosy+cos²y=1/16sinx-siny=1/3sin²x-2sinxsiny+sin²y=1/9两式相
对x求导,这里y是关于x的函数,所以有y'=(cosx)'+(1/2siny)'=-sinx+1/2cosy*y'整理得y'=2sinx/(cosy-2)