sin+x+y+展开式
甫文虾2330化简sin(x+y)sin(x - y)+cos(x+y)cos(x - y)除了展开,有简便算法么? -
苍向狠15530976923 ______ 这个很简单啊 cos(a-β)=cosacosβ+sinasinβ 这里就是这样啊,把x+y=a,x-y=β带入 sin(x+y)sin(x-y)+cos(x+y)cos(x-y)=cos[(x+y)-(x-y)]=cos2y
甫文虾2330sin(x+y)+y=0隐函数求导 -
苍向狠15530976923 ______ 两边同时对x求导-(1+y')sin(x+y)=cos(x-y)(1-y')所以y'=【cos(x-y)+sin(x+y)】/【cos(x-y)-sin(x+y)】
甫文虾2330cos(X - Y)=0.5 sin2X+sin2Y=2/3 则sin(X+Y)=---------- -
苍向狠15530976923 ______ sin2X+sin2Y=2/3,2sin(x+y)cos(x-y)=2/3,将cos(X-Y)=0.5代入,就可以了.用的是【和差化积】公式.注x,y与X,Y我没有区分开.
甫文虾2330sinx的立方如何展开?? -
苍向狠15530976923 ______ 用三倍角公式,sin(3α) = 3sinα-4sin^3α (sinx)^3=3/4sinx-1/4sin(3x) =3/4(x-1/3!*x^3+1/5!*x^5-1/7!*x^7+...)-1/4(3x-1/3!*(3x)^3+1/5!*(3x)^5-1/7!*(3x)^7+...) =3/4Σ(-1)^n*1/(2n+1)!*x^(2n+1)-1/4Σ(-1)^n*1/(2n+1)!*(3x)^(2n+1) =Σ(-1)^n*1/(2n+1)!*[3/4-1/4*3^(2n+1)]x^(2n+1) n从0到无穷大
甫文虾2330证明sin(x+y)=sinx*cosy+cosx*siny的过程 -
苍向狠15530976923 ______ 推荐一种向量法证明:在单位圆上取两点M N,与x轴的夹角分别是x,pi/2+y 则M(cosx,sinx),N(-siny,cosy) (OM,ON)= cos(OM,ON)=cos(pi/2+y-x)=sin(x-y)=OM点乘ON/(|OM||ON|) 即 sin(x-y)=sinxcosy-cosxsiny 用-y替换y可得:sin(x+y)=sinxcosy+cosxsiny 证毕 类似的可得到cos(x+y)的公式
甫文虾2330化简:sin(x+y)cosx - 1/2[sin(2x+y) - siny] -
苍向狠15530976923 ______[答案] sin(2x+y)=sin(x+y+x)=sin(x+y)*cosx+cos(x+y)*sinxsiny=sin(x+y-x)=sin(x+y)*cosx-cos(x+y)*sinx所以,sin(x+y)cosx-1/2[sin(2x+y)-siny]=sin(x+y)*cosx-cos(x+y)*sinx=sin(x+y-x)=siny
甫文虾2330sin(x+y+1)=dy/dx化为可分离变量方程,并求解 -
苍向狠15530976923 ______ 令x+y+1=u y=u-x-1 dy/dx=du/dx-1 那么sinu=du/dx-1 du/dx=sinu+1 分离变量得 du/(sinu+1)=dx du/[sin(u/2)+cos(u/2)]²=dx du/{cos²(u/2)[tan(u/2)+1]²}=dx 2d[tan(u/2)+1]/[tan(u/2)+1]²=dx -2/[tan(u/2)+1]=x+C -2/[tan(x+y+1)/2+1]=x+C
甫文虾2330若等式sinx+siny=sin(x+y)成立,则必有( ) -
苍向狠15530976923 ______[选项] A. x∈R,y∈R B. x=y=nπ,(n∈Z) C. x=-y D. x,y,x+y中,至少有一个为2nπ(n∈Z)
甫文虾2330求三角函数的周期,有一个 好像叫叠加公式的,就是 y=根号sin(..x+&) 这个式子写完整是什么样? -
苍向狠15530976923 ______[答案] y=Asin(ωx+ φ)+B
甫文虾2330若等式sinx+siny=sin(x+y)成立,则必有( ) -
苍向狠15530976923 ______[选项] A. x∈R,y∈R B. x=y=nπ,(n∈Z) C. x=-y D. x,y,x+y中,至少有一个为2nπ(n∈Z)