sin+x-y
答：sin(x-y)=sinx cosy - cosx siny
答：sin(α±β)=sinα·cosβ±cosα·sinβ tan(α+β)=(tanα+tanβ)/(1-tanα·tanβ)tan(α-β)=(tanα-tanβ)/(1+tanα·tanβ)
答：一定是。因为sinx是奇函数 f(-x)=-f(x)奇函数的代数意义是：当自变量互为相反数时，函数值也互为相反数。奇函数的几何意义是：图象关于原点对称。f(x)=sinx f(x-y)=f(-(y-x))=-f(y-x)即sin(x-y)=sin(-(y-x))=-sin(y-x)f(xy)=f(-(-xy))=-f(-xy)即……...
答：解：∵y'=sin(x-y)==>dy/dx=sin(x-y)==>1-(1-dy/dx)=sin(x-y)==>1-d(x-y)/dx=sin(x-y)==>d(x-y)/dx=1-sin(x-y)==>d(x-y)/(1-sin(x-y))=dx ==>(1+sin(x-y))d(x-y)/((1-sin(x-y))(1+sin(x-y)))=dx (等式左端分子分母同乘(1+sin(x-y)...
答：作变量代换，令x－y＝u，则dy/dx＝1－(du/dx)，代入原方程，可得du/dx＝1－sinu，即将原方程转化为可分离变量方程，求解上述方程即可。
答：y=sin（x-y）y'=cos（x-y）*（1-y'）y'+y'*cos（x-y）=cos（x-y）y'=cos（x-y）/[1+cos（x-y）]
答：sin(x-y)=u y=x-arcsinu dy/dx=1-[1/√(1-u^2)]du/dx 1-1/[√(1-u^2)]du/dx=u (1-u)=[1/√(1-u^2)]du/dx dx=du/[(1-u)√(1-u^2)]∫du/[(1-u)√(1-u^2)]设u=cosv 1+u=1+cosv=2[cos(v/2)]^2 √(1-u^2)=sinv=2sinv/2cosv/2 ...
答：简单分析一下即可，答案如图所示
答：一般含绝对值的函数都采用零点分区间法，就是令绝对值里等于零做出数轴然后再分区间去绝对值，也就变成分段函数。|sin(x-y)|去绝对值要考虑|sin(x-y)|＜0和|sin(x-y)|＞0，令x-y＝x，作出y=Sinx的图像，把零点找出，分区间写出y＞0和y＜0的x取值范围，最后用x-y代入x即可求解同时还要...
答：==>d(x-y)/(sin((x-y)/2)-cos((x-y)/2))^2=dx (应用(sin((x-y)/2)^2+(cos((x-y)/2)^2=1)==>(sec((x-y)/2))^2d(x-y)/(tan((x-y)/2)-1)^2=dx (分子分母同除(cos((x-y)/2))^2)==>2d(tan((x-y)/2)-1)/(tan((x-y)/2)-1)^2=dx =...
网友评论：
翁泉17266241414：
想问一下 sinx加siny等于sin(x加y)吗 这里两个都是sin -
2109越哈
： 不等,sin是正弦函数的符号. sin(x+y)=sinxcosy+cosxsiny sinx+siny=2sin[(x+y)/2]cos[(x-y)/2]
翁泉17266241414：
急:sin(x+y)=sinxcosy+cosxsiny怎么证明??? -
2109越哈
： 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
翁泉17266241414：
sin(x+y)sin(x - y)=sin^2x - sin^2y -
2109越哈
： 本题应用积化和差公式:(先说说公式推导:cos(x+y)=cosxcosy-sinxsiny; cos(x-y)=cosxcosy+sinxsiny; 于是sinxsiny=(cos(x-y)-cos(x+y))/2;) 将公式应用于本题 左边=sin(x+y)sin(x-y)=(cos2y-cos2x)/2 =(1-2sin^2y-1+2sin^2x)/2 =sin^2x-sin^2y=右边 所以命题得证;
翁泉17266241414：
y'+sin(x+y)=sin(x - y)的通解 -
2109越哈
：[答案] y'=sin(x-y)-sin(x+y)=-2cosxsiny, ∴dy/siny=-2cosxdx, 积分得lntan(y/2)=-2sinx+C', tan(y/2)=Ce^(-2sinx), y/2=kπ+arctan[Ce^(-2sinx)],k∈Z, y=2kπ+2arctan[Ce^(-2sinx)],为所求.
翁泉17266241414：
证明sin(x+y)sin(x - y)=sinx - siny -
2109越哈
： 解:sin(x+y)sin(x-y) =-1/2(cos(x+y+x-y)—cos(x+y-x+y)) =-1/2(cos2x—cos2y) =-1/2(1-2(sinx)^2-1+2(siny)^2) =(sinx)^2-(siny)^2
翁泉17266241414：
设sin(x+y)sin(x - y)=m,则cos^2x - cos^2y的值 -
2109越哈
： sin(x+y)sin(x-y)=[sinxcosy+sinycosx][sinxcosy-cosxsiny]=(sinxcosy)^2-(cosxsiny)^2=(1-cos^2y)cos^2y-cos^2x(1-cos^2y)=cos^2y-cos^2x=m 故cos^2x-cos^2y=-m
翁泉17266241414：
为什么sin(x+y) - sinx = 2cos(x+y/2)siny/2 -
2109越哈
： 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
翁泉17266241414：
化简sin(x+y)cos(x - y)+cos(x+y)sin(x - y) -
2109越哈
：[答案] sin(x+y)cos(x-y)+cos(x+y)sin(x-y) =sin(x+y+x-y) =sin2x
翁泉17266241414：
三角函数sin(x - y)cosy+cos(x - y)siny化简要详细过程
2109越哈
： =(sinxcosy-cosxsiny)cosy+(cosxcosy+sinxsiny)siny =sinxcos^2y-cosxsinycosy+cosxcosysiny+sinxsin^2y =sinx(cos^2y+sin^2y) 因为sin^2x+cos^2y=1 所以最后=sinx
翁泉17266241414：
sin(x+y) - sinx= -
2109越哈
： sin(x+y)-sinx=2sin[(x+y)-x]/2*cos[(x+y)+x]/2=2sin(y/2)cos[(2x+y)/2]希望对您有所帮助 如有问题,可以追问. 谢谢您的采纳
