求证:sin(2x-y) sinx-2cos(x-y)=-siny sinx

sin(x+y)=1所以x+y=2kπ+π/2所以2x+y=2(x+y)-y=4kπ+π-y所以tan(2x+y)=tan(4kπ+π-y)因为tan的周期是π所以tan(4kπ+π-y)=tan[4

(x/a)cosθ+(y/b)sinθ=1[(x/a)cosθ+(y/b)sinθ]^2=1(x/a)sinθ-(y/b)cosθ=1[(x/a)sinθ-(y/b)cosθ]^2=1[(x/a)co

合并同类项么,很简单的只要你愿意去做左边=cos*x（cos*y+sin*y）+sin*x（cos*y+sin*y)=cos*x+sin*x=1=右边

左边=(sinxcosy+cosxsiny)(sinxcosy-cosxsiny)=sin²xcos²y-cos²xsin²y=sin²x(1-sin

你看后面TAN里一个x一个x+y那你就把给你的原式中的2x+y拆开,在消消化化的,试下吧我觉得能行

令a=x+y,则条件变为3sin(a-x)=sin(a+x),展开得3sinacosx-3cosasinx=sinacosx+cosasinx,移项2sinacosx=4cosasinxtana=2t

dy/dx相当于对x进行求导：dy/dx=y'=2x*cos[sin(x^2)]*cos(x^2)由于：sinx=cosx,sin(x^2)=2x*cos(x^2)

已知sin（x+y）=1,求证：tan（2x+3y）=tany证明：sin（x+y）=1所以x+y=2k兀+兀/2K为整数所以tan(2x+3y)=tan(4k兀+兀+y)=tan(兀+y)=tany

sinx+siny+sinz-sin(x+y+z)=4sin[(x+y)/2]sin[(x+z)/2]sin[(y+z)/2]sinx+siny+sinz-sin(x+y+z)=2sin[(x+y)/

sin[(x+y)+x]=5sin[(x+y)-x]sin(x+y)·cosx＋cos(x+y)·sinx＝5·sin(x+y)·cosx－5·cos(x+y)·sinx4·sin(x+y)·cosx

x=(x+y)/2+(x-y)/2y=(x+y)/2-(x-y)/2所以左边=cos[(x+y)/2+(x-y)/2]-cos[(x+y)/2-(x-y)/2]={cos[(x+y)/2]cos[(x

sinx+siny=sin[(x+y)/2+(x-y)/2]+sin[(x+y)/2-(x-y)/2]=sin(x+y)/2*cos(x-y)/2+cos(x+y)/2*sin(x-y)/2+sin(

|2sin[(x-y)/2]cos[(x+y)/2]|=I2[sin(x/2)cos(y/2)-sin(y/2)cos(x/2)][cos(x/2)cos(y/2)-sin(x/2)(sin(y/2)

把5siny=sin(2x+y)变为5sin[(x+y)-x]=sin[（x+y)+x],把其中的（x+y）,看成一个整体,上式即变为4sin(x+y)cosx=6cos(x+y)sinx,再把式子的

你可以把分母sinx-siny用和差化积化成2sin((x-y)/2)cos((x+y)/2)这样答案就很显然了

由sin(x+y)=1可知x+y=90度tan(2x+y)+tanytan(2x+y+y)tan(2x+2y)tan180度因为:tan180度=0(常识!)所以:tan(2x+y)+tany=0

sin(x-y)=sinxcosy-cosxsiny,sin(x+y)=sinxcosy+cosxsinysin(x-y)sin(x+y)=sin²xcos²y-cos²

前三题其实就是和差化积的公式,4因为tan2a=2tana/(1-tan^2a)sin2a=2tana/(1+tan^2a)所以左边=2tana/(1+tan^2a)-√3cos2a.先消去一个tan

sin^2x+sin^2y-sin^2x*sin^2y+cos^2x*cos^2y=sin^2x-sin^2x*sin^2y+sin^2y+cos^2x*cos^2y=sin^2x*(1-sin^2y

我来给你解答,稍等再答：