2x^2 y^2-cosy=0所确定的隐函数

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

极值就是求导fx=cosx-sin(x-y)=0fy=-siny+sin(x-y)=0x+y=pi/2f(x,y)=1+0+0=1极小值这是f(x,y)

解cosx+cosy-cos(x+y)=3/22cos[(x+y)/2]*cos[(x-y)/2]-2{cos[(x+y)/2]}^2=1/2{cos[(x-y)/2]-2cos[(x+y)/2]}^

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

3:通分,将3个分式的分母化成相同的,即sin(x-y)sinz/(sinxsinysinz)+sin(y-z)sinx/(sinxsinysinz)+sin(z-x)siny/(sinxsinysi

∵(1+y²)dx+(xy-√(1+y²)cosy)dy=0==>√(1+y²)dx+(xy/√(1+y²)-cosy)dy=0(等式两端同除√(1+y

思路：利用极值和导数的关系（极值点,导数为0）函数关于x,y求偏导数,令其为0,解出x,y的值,和相应的函数值,那就是极值

cos(2x-2y)=cos2(x-y)=2cos^2(x-y)-1cos(x-y)=cosxcosy+sinxsiny(sinx+siny)^2=sin^2x+sin^2y+2sinxsiny=4/

把cosy看作新的因变量,令z＝cosy,原方程化为dz/dx＋2/x×z＝-3,一个线性方程,套用通解公式,z＝1/x^2×(-x^3＋C).原方程的通解是cosy＝1/x^2×(-x^3＋C),即

cos(2x+y)=cos[(x+y)+x]=cos(x+y)cosx-sin(x+y)sinxcosy=cos[(x+y)-x]=cos(x+y)cosx+sin(x+y)sinx3cos（2x+y

消元sinx=1-siny----（1cosx=-cosy-----（2(1)式^2+(2)式^2--------1=(1-siny)^2+cosy^21=1+sin^2y+cos^2y-2siny1

∫∫(D)cosy²dxdy=∫(0,2)cosy²dy∫(1,y+1)dx(∫(a,b)表示从a到b积分)=∫(0,2)cosy²*[(y+1)-1]dy=∫(0,2)

x=(x+y)/2+(x-y)/2y=(x+y)/2-(x-y)/2所以左边=cos[(x+y)/2+(x-y)/2]-cos[(x+y)/2-(x-y)/2]={cos[(x+y)/2]cos[(x

题目有没有问题呢?怎么觉得应该是求tan（x＋y）tanx呢?∵2cos（2x＋y）＝cosy∴2cos（x＋x＋y）＝cos（x＋y－x）∴2cosxcos(x＋y)－2sinxsin(x＋y)＝c

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

两式分别求平方得：sinx2-2sinxsiny+siny2=1/4,cosx2-2cosxcosy+cosy2=1/4相加得：2-2（sinxsiny+cosxcosy）=1/2,sinxsiny+

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

y`=x^2/cosysinydy=x^2dx积分得通-cosy=x^3/3+C或：y=arccos(-x^3/3-C)再问：sinydy=x^2dx，拜托--你都cosy，变成siny了还dy啊再答

cos(2x+y)=3cosycos(x+y+x)=3cos(x+y-x)cos(x+y)cosx-sin(x+y)sinx=3[cos(x+y)cosx+sin(x+y)sinx]2cos(x+y)