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cosxcosy的积分

来源:baiyundou.net   日期:2024-09-29

武狗琪4358cosx - Cosy等于什么 -
阚卓苗17846102140 ______ =cos²x-cos²y =(1+cos2x)/2-(1+cos2y)/2 =(cos2x-cos2y)/2 ={cos[(x+y)+(x-y)]-cos[(x+y)-(x-y)]}/2 =[cos(x+y)cos(x+y)-sin(x+y)sin(x-y)-cos(x+y)cos(x+y)-sin(x+y)sin(x-y)]/2 =-sin(x+y)sin(x-y)

武狗琪4358sinθ=y/r,cosθ=x/r, tanθ=y/x, cotθ=x/y怎么证明? -
阚卓苗17846102140 ______ 坐标系内,画个原点o为圆心,r为半径的圆!第一象限内任意取圆上一点p(x,y) 连接op 做pD垂直于x轴于D,记<poD=θ 则:sinθ=pD/r=y/r cosθ=x/r, tanθ=y/x, cotθ=x/y

武狗琪4358已知cosX+cosY=3/5,sinX+sinY=4/5,求cos(X+Y)的值 -
阚卓苗17846102140 ______ (cosx+cosy)^2=9/251+2cosxcosy=9/25 cosxcosy= -8/25 同理 sinxsiny= -1/10 cos(x+y)=cosxcosy-sinxsiny=-16/50+5/50= -11/50

武狗琪4358求微分方程满足初始条件的特解 -
阚卓苗17846102140 ______ sinydy/cosy=sinxdx/cosx 两边同时积分 - ∫1/cosydcosy= - ∫1/cosxdcosx l...

武狗琪4358siny*cosx*dy=cosy*sinx*dx的通解,这题怎么做? -
阚卓苗17846102140 ______ 显然,cosy=0,即y=2kπ±π/2是原方程的特解,其中k是任意整数 当cosy≠0时 tanydy=tanxdx ∫tanydy=∫tanxdx ln|secy|=ln|secx|+C secy=C*secx cosy=C*cosx y=2kπ±arccos(C*cosx),其中C是任意非零常数 综上所述,y=2kπ±arccos(C*cosx),其中C是任意常数,k是任意整数

武狗琪4358matlaB:、解微分方程cosxsinydy=cosysinxdx,x=0时,y=4/pi -
阚卓苗17846102140 ______[答案] ∵cosxsinydy=cosysinxdx ==>sinydy/cosy=sinxdx/cosx ==>d(cosy)/cosy=d(cosx)/cosx ==>ln│cosy│=ln│cosx│+ln│C│ (C是积分常数) ==>cosy=C*cosx ∴原方程的通解是cosy=C*cosx (C是积分常数) ∵当x=0时,y=π/4 ∴cos(π/4)=C*cos(0) ==>C=1/...

武狗琪4358y'=cos(x - y) - cos(x+y)的通解,求详解. -
阚卓苗17846102140 ______[答案] ∵y'=cos(x-y)-cos(x+y) ==>y'=(cosxcosy+sinxsiny)-(cosxcosy-sinxsiny) (应用余弦和差角公式) ==>y'=2sinxsiny ==>dy/siny=sinxdx ==>∫[1/(cosy-1)-1/(cosy+1)]d(cosy)=2∫sinxdx ==>ln│(cosy-1)/(cosy+1)│=-2cosx+ln│C│ (C是积分常数) ==>(...

武狗琪4358求sinxcosy的不定积分 -
阚卓苗17846102140 ______[答案] sinxcosx=0.5sin2x 积分为-0.25cos2x

武狗琪4358如何求cos3xcos2xdx的不定积分书上讲解的是把他化解为了(cos5x+cosx)/2后求的积分.我不明白怎么把cos3xcos2x化成的(cos5x+cosx)/2. -
阚卓苗17846102140 ______[答案] COSx*COSy=[COS(x+y)+COS(x-y)]/2

武狗琪4358y'=cos(x - y) - cos(x+y)的通解,求详解. -
阚卓苗17846102140 ______ 解:∵y'=cos(x-y)-cos(x+y) ==>y'=(cosxcosy+sinxsiny)-(cosxcosy-sinxsiny) (应用余弦和差角公式) ==>y'=2sinxsiny ==>dy/siny=sinxdx ==>∫[1/(cosy-1)-1/(cosy+1)]d(cosy)=2∫sinxdx ==>ln│(cosy-1)/(cosy+1)│=-2cosx+ln│C│ (C是积分常数) =...

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