cis+x+y
@荆马5380：为什么sin(x+y) - sinx = 2cos(x+y/2)siny/2 -
闫图18651781086…… x+y=x+y/2+y/2 x=x+y/2-y/2 sin(x+y)=sin(x+y/2+y/2)=sin(x+y/2)cosy/2+cos(x+y/2)siny/2 sinx=sin(x+y/2)cosy/2-cos(x+y/2)siny/2 所以 sin(x+y) - sinx =sin(x+y/2)cosy/2+cos(x+y/2)siny/2-[sin(x+y/2)cosy/2-cos(x+y/2)siny/2]= 2cos(x+y/2)siny/2
@荆马5380：急:sin(x+y)=sinxcosy+cosxsiny怎么证明??? -
闫图18651781086…… 用三角函数的积化和差公式,sinXcosY=[sin(X+Y)+sin(X-Y)]/2 cosXsinY=[sin(X+Y)-sin(X-Y)]/2 两个式子一加,消去sin(X-Y)即可
@荆马5380：sin(x+y)sin(x - y) - cos(x+y)cos(x - y)的详细解法
闫图18651781086…… sin(x+y)sin(x-y)-cos(x+y)cos(x-y)=-[cos(x+y)cos(x-y)-sin(x+y)sin(x-y)] =-[cos((x+y)-(x-y))]=-cos2y
@荆马5380：【高中数学】已知cos(x+y)=1/3,cos(x - y)=2/3,且0 - 作业帮
闫图18651781086…… [答案] 0π/3-π/3所以sin(x+y)=2根号2/3,sin(x-y)=-根号5/3 所以 cos2x =cos(x+x+y-y) =cos(x+y+x-y) =cos(x+y)cos(x-y)-sin(x+y)sin(x-y) =2/9+2根号10/9 =2/9(1-根号10)
@荆马5380：计算二重积分:xsin(x+y)dσ , D:0≤x≤π,0≤y≤π/2 -
闫图18651781086…… 两重积分先积sin(x+y)dy II xsin(x+y)dσ=Ixdx Isin(x+y)dy =I xdx I sin(x+y)d(x+y) =I xdx(-cos(x+y)):pi/2-0 =I xdx(-cos(x+pi/2)+cos(x)) =I xcos(x)dx- I xcos(x+pi./2)dx =(xsinx- I sinxdx) - I (x+pi/2)cos(x+pi/2)dx+pi/2 * I cos(x+pi/2)dx =cosx -(x+pi/2)sin(x+pi/2)-cos...
@荆马5380：求sin(x+y)的二重积分,x,y均[0,pai/2] -
闫图18651781086…… 直线x+y=π将积分区间分成2部分,左下部分记为D1,右上部分记为D2 则先积D1里的积分,在D1内,由于x+yπ] dx∫[0--->π-x] sin(x+y) dy =-∫[0--->π] cos(x+y) |[0--->π-x] dx =∫[0--->π] (cosx-cosπ) dx =∫[0--->π] (cosx+1) dx =sinx+x |[0--->π] =π 对于D2,ππ]...
@荆马5380：函数y=cis²X+4的奇偶性 -
闫图18651781086…… cos...吧 f(x)=cos²X+4 f(-x)=cos ²(-x)+4 又∵cosx=cos(-x) ∴ f(x)= f(-x)(x∈R) ∴为偶函数
@荆马5380：求函数z=arcsin(x+y)的全微积分 -
闫图18651781086…… ∂z/∂x=(x+y)'/√[1-(x+y)²]=1/√[1-(x+y)²] ∂z/∂y=(x+y)'/√[1-(x+y)²]=1/√[1-(x+y)²] 所以 dz=dx/√[1-(x+y)²] + dy/√[1-(x+y)²] =(dx+dy)/√[1-(x+y)²]
@荆马5380：cos(x)+cos(x+y)+cos(x+2y)+.....+cos(x+ny) -
闫图18651781086…… 不是不可以做,只是计算太繁琐了. 令z=cosx+isinx,w=cosy+isiny 则原式 =z+zw+zw^2+...+zw^(n-1) =z(w^n-1)/(w-1) =(cosx+isins)(cosny+isinny-1)/(cosy+isiny-1) 得到以上式子之后就把它化成A+iB的形式,那么原式就等于A了 像这种正弦或余弦求和,一般是用复数来做,屡试不爽,只是计算繁琐
@荆马5380：已有条件 x=cosθ+isinθ (即:cisθ) 求证 x^n+1/(x^n)=2cis(nθ -
闫图18651781086…… x^n=cosnθ+isinnθ1/x^n=cosnθ-isinnθ x^n+1/x^n=2cosnθ
闫图18651781086…… x+y=x+y/2+y/2 x=x+y/2-y/2 sin(x+y)=sin(x+y/2+y/2)=sin(x+y/2)cosy/2+cos(x+y/2)siny/2 sinx=sin(x+y/2)cosy/2-cos(x+y/2)siny/2 所以 sin(x+y) - sinx =sin(x+y/2)cosy/2+cos(x+y/2)siny/2-[sin(x+y/2)cosy/2-cos(x+y/2)siny/2]= 2cos(x+y/2)siny/2
@荆马5380：急:sin(x+y)=sinxcosy+cosxsiny怎么证明??? -
闫图18651781086…… 用三角函数的积化和差公式,sinXcosY=[sin(X+Y)+sin(X-Y)]/2 cosXsinY=[sin(X+Y)-sin(X-Y)]/2 两个式子一加,消去sin(X-Y)即可
@荆马5380：sin(x+y)sin(x - y) - cos(x+y)cos(x - y)的详细解法
闫图18651781086…… sin(x+y)sin(x-y)-cos(x+y)cos(x-y)=-[cos(x+y)cos(x-y)-sin(x+y)sin(x-y)] =-[cos((x+y)-(x-y))]=-cos2y
@荆马5380：【高中数学】已知cos(x+y)=1/3,cos(x - y)=2/3,且0 - 作业帮
闫图18651781086…… [答案] 0π/3-π/3所以sin(x+y)=2根号2/3,sin(x-y)=-根号5/3 所以 cos2x =cos(x+x+y-y) =cos(x+y+x-y) =cos(x+y)cos(x-y)-sin(x+y)sin(x-y) =2/9+2根号10/9 =2/9(1-根号10)
@荆马5380：计算二重积分:xsin(x+y)dσ , D:0≤x≤π,0≤y≤π/2 -
闫图18651781086…… 两重积分先积sin(x+y)dy II xsin(x+y)dσ=Ixdx Isin(x+y)dy =I xdx I sin(x+y)d(x+y) =I xdx(-cos(x+y)):pi/2-0 =I xdx(-cos(x+pi/2)+cos(x)) =I xcos(x)dx- I xcos(x+pi./2)dx =(xsinx- I sinxdx) - I (x+pi/2)cos(x+pi/2)dx+pi/2 * I cos(x+pi/2)dx =cosx -(x+pi/2)sin(x+pi/2)-cos...
@荆马5380：求sin(x+y)的二重积分,x,y均[0,pai/2] -
闫图18651781086…… 直线x+y=π将积分区间分成2部分,左下部分记为D1,右上部分记为D2 则先积D1里的积分,在D1内,由于x+yπ] dx∫[0--->π-x] sin(x+y) dy =-∫[0--->π] cos(x+y) |[0--->π-x] dx =∫[0--->π] (cosx-cosπ) dx =∫[0--->π] (cosx+1) dx =sinx+x |[0--->π] =π 对于D2,ππ]...
@荆马5380：函数y=cis²X+4的奇偶性 -
闫图18651781086…… cos...吧 f(x)=cos²X+4 f(-x)=cos ²(-x)+4 又∵cosx=cos(-x) ∴ f(x)= f(-x)(x∈R) ∴为偶函数
@荆马5380：求函数z=arcsin(x+y)的全微积分 -
闫图18651781086…… ∂z/∂x=(x+y)'/√[1-(x+y)²]=1/√[1-(x+y)²] ∂z/∂y=(x+y)'/√[1-(x+y)²]=1/√[1-(x+y)²] 所以 dz=dx/√[1-(x+y)²] + dy/√[1-(x+y)²] =(dx+dy)/√[1-(x+y)²]
@荆马5380：cos(x)+cos(x+y)+cos(x+2y)+.....+cos(x+ny) -
闫图18651781086…… 不是不可以做,只是计算太繁琐了. 令z=cosx+isinx,w=cosy+isiny 则原式 =z+zw+zw^2+...+zw^(n-1) =z(w^n-1)/(w-1) =(cosx+isins)(cosny+isinny-1)/(cosy+isiny-1) 得到以上式子之后就把它化成A+iB的形式,那么原式就等于A了 像这种正弦或余弦求和,一般是用复数来做,屡试不爽,只是计算繁琐
@荆马5380：已有条件 x=cosθ+isinθ (即:cisθ) 求证 x^n+1/(x^n)=2cis(nθ -
闫图18651781086…… x^n=cosnθ+isinnθ1/x^n=cosnθ-isinnθ x^n+1/x^n=2cosnθ
