cos+x+y
段姣齿3451已知x,y均为锐角,cos(x+y)=12/13,cos(2x+y)=3/5,求cos x 需要解题过程,谢谢 -
荀刘泉17886892872 ______ 00所以0所以sin(x+y)>0 sin²(x+y)+cos²(x+y)=1 所以sin(x+y)=5/1300所以0<2x+y<270 cos(2x+y)=3/5>0 则只有0<2x+y<90 所以sin(2x+y)>0 sin²(2x+y)+cos²(2x+y)=1 所以sin(x+y)=4/5 所以cosx=cos[(2x+y)-(x+y)]=cos(2x+y)cos(x+y)+sin(2x+y)sin(x+y)=56/65
段姣齿3451已知y=cos(x+y),求dy -
荀刘泉17886892872 ______ dy=d(cos(x+y)) dy=-sin(x+y)d(x+y) dy+sin(x+y)(dx+dy)=0 (1+sin(x+y))dy=-sin(x+y)dx dy=-sin(x+y)dx/(1+sin(x+y))
段姣齿3451cosxy=x的导数和y=cos(x+y)的导数怎么求,详细过程,谢谢! -
荀刘泉17886892872 ______ cosxy=x -sinxy*(y+xy')=1 y+xy'=-cscxy y'=-(cscxy+y)/x. y=cos(x+y) y'=-sin(x+y)*(1+y') y'[1+sin(x+y)]=-sin(x+y) y'=-sin(x+y)/[1+sin(x+y)].
段姣齿3451y=cos(x+y),求y'速度啊 -
荀刘泉17886892872 ______ y'=-(x+y)'sin(x+y) =-(1+y')sin(x+y) 所以, y'[1+sin(x+y)]=-sin(x+y) 即, y'=-sin(x+y)/[1+sin(x+y)]
段姣齿3451cosx+cosy=1/2,sinx - siny=1/3,则cos(x+y)=? -
荀刘泉17886892872 ______ 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=13/36 由cos^2x+si...
段姣齿3451已知x,y∈(0,π),求满足方程cosX+cosy - cos(x+y)=3/2的x,y的值 求过程 -
荀刘泉17886892872 ______ cosx+cosy=2cos((x+y)/2)*cos((x-y)/2),cos(x+y)=2[cos((x+y)/2)]^2-1 cosX+cosy-cos(x+y)=3/2可以变为2cos((x+y)/2)*cos((x-y)/2)-2[cos((x+y)/2)]^2+1=3/2 整理为4[cos((x+y)/2)]^2-4cos((x+y)/2)*cos((x-y)/2)+1=0① 由△≥0,得16[cos((x-y)/2)]^2-16≥0,即[...
段姣齿34512cos(x+y)cosx - cos(2x+y)
荀刘泉17886892872 ______ 解:2cos(x+y)cosx-cos(2x+y) =2cos(x+y)cosx-cos[x+(x+y)] =2cos(x+y)cosx-[cos(x+y)cosx-sin(x+y)sinx] =cos(x+y)cosx+sin(x+y)sinx =cos(x+y). 抱歉,结果我写快了,写错了,应该是:cos(x+y-x)=cosy. 解:2cos(x+y)cosx-cos(2x+y) =2cos(x+y)cosx-cos[x+(x+y)] =2cos(x+y)cosx-[cos(x+y)cosx-sin(x+y)sinx] =cos(x+y)cosx+sin(x+y)sinx =cos(x+y-x) =cosy.
段姣齿3451求证COS(x+y)COS(x - y)=COS2x - SIN2y
荀刘泉17886892872 ______ zhengming: cos(x+y)=cosxcosy-sinxsiny cos(x-y)=cosxcosy+sinxsiny所以,二者相乘以后=cosxcoxcosycosy-sinxinxsinysiny=(1-sinxsinx)cosycosy-sinxsinxsinysiny=cos2x-sin2y
段姣齿3451y=sinxy+cos(x+y)求y对x的偏导数 -
荀刘泉17886892872 ______[答案] y`=cosxy(y+xy`)-sin(x+y)(1+y`) =ycosxy+xy`cosxy-sin(x+y) -y`sin(x+y) y`-xy`cosxy+y`sin(x+y)=ycosxy-sin(x+y) y`=[ycosxy-sin(x+y)]/[1-xcosxy+sin(x+y)]
段姣齿3451求z=xcos(x+y)的偏导数 -
荀刘泉17886892872 ______[答案] 对x求导时,y看成常数: z'(x)=cos(x+y)-xsin(x+y)——这用到积求导公式 对y求导时,x看成常数: z'(y)=-xsin(x+y)