cosxsinydy=cosysinxdx y|x=0=
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∵cos(x+y)cosy+sin(x+y)siny=0==>cos[(x+y)-y]=0(应用余弦差角公式cos(A-B)=cosAcosB+sinAsinB)==>cosx=0∴cosx=0.
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
两边求Ln,得到cosy*ln(sinx)=siny*ln(cosx),化简得y=acrtan(lnsinx/lncosx)公式:y=arctanxy'=1/1+x^2带入上面y'=(1/1+(lns
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(cosxsiny)dx+(sinxcosy)dy=0sinydsinx+sinxdsiny=0dsinx/sinx+dsiny/siny=0d(lnsinx)+d(lnsiny)=0d(ln(sin
题目有问题再问:已知sinx+cosy=3/4,cosx+cosy=1/2,求tanx*tany,题目打错了,不用和差化积公式怎么做?再答:你只要求出cosy为多少即可以,只要把俩个等式的COSY都移
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
1)12)1再问:第二题不是4?再答:不是再问:求解释再答:用双角公式:cos(x)=1-2sin(x/2)^2再问:求过程,不懂怎么算出来的,麻烦了再答:cos(x)=1-2sin(x/2)^20.
∵cosxsinydy=cosysinxdx==>sinydy/cosy=sinxdx/cosx==>d(cosy)/cosy=d(cosx)/cosx==>ln│cosy│=ln│cosx│+ln│
(sinx)^y=(cosy)^x两边取对数ln(sinx)^y=ln(cosy)^xyln(sinx)=xln(cosy)两边求导:y'ln(sinx)y/sinx*cosx=ln(cosy)x/c
ezplot('y-sin(x)*cos(y)')再问:是的,不过用mathematica画出来是个三维图形啊,你这个是一条线对吗?再答:嗯,MATLAB里画出来是一条线。MATHEM
第一题:设t=cosx+cosy②sinx+siny=1①两式平方再相加t^+1=2+2sinxsiny+2cosxcosyt^=2cos(x-y)+1-1≤t^≤3既0≤t^≤3-根号3≤cosx+
sinx-siny=-2/3cosx+cosy=2/3平方得sin^2x+sin^2y-2sinxsiny=4/9cos^2x+cos^2y+2cosxcosy=4/9相加得2+2(cosxcosy-
如果xy=sinxcosy,那么x=sinxcosy/ydy/dx=dxy/dx*2=dsinxcosy/dx*2=dsinxcosy/d(sinx*2cosy*2/y*2)=y这里的*是平方的意思
原式=cosacosy+sinasiny=cosa(cosa+cosb)+sina(sina+sinb)=1+cosacosb+sinasinbsin^2(y)+cos^2(y)=1+1+2sinas
sinx+siny=2sin[(x+y)/2]cos[(x-y)/2]sinx-siny=2cos[(x+y)/2]sin[(x-y)/2]cosx+cosy=2cos[(x+y)/2]cos[(x-
令sinx+siny=α,给两式子都平方再相加:1+α²=2sinxsiny+2cosycosx+cos²x+sin²x+cos²y+sin²y1+α
sinx=1/3-siny故sinx-(cosy)^2=(1/3-siny)-[1-(siny)^2]=(siny)^2-siny-2/3=(siny-1/2)^2-1/4-2/3siny∈[-1,1
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
y'cosy=x-siny;设p=siny;p'+p=x;Pe^x=xe^x-e^x+C