x+siny和y+arcsinx的区别

扶炎心2331化简cos(x+y)sinx - sin(x+y)cosx -
侯雄发13236529164 ______ cos(x+y)sinx-sin(x+y)cosx =-[sin(x+y)cosx-cos(x+y)sinx]=-sin(x+y-x)=-siny 参考公式: sin(α-β)=sinαcosβ-cosαsinβ

扶炎心2331求极限,x趋向于0时,求sinx/x+arc sinx/x+tanx/x+arc tanx/x的极限 -
侯雄发13236529164 ______[答案] x→0,则sinx~arcsinx~tanx 【它们之间在x→0下为等价无穷小】∴lim(x→0)(sinx/x+arc sinx/x+tanx/x+arc tanx/x)=lim(x→0)(sinx/x)+lim(x→0)(arcsinx/x)+lim(x→0)(tanx/x)+lim(x→0)(arctanx/x)=lim(x→0)(x/x...

扶炎心2331化简sinxsin(x+y)+cosxcos(x+y) ,答案里面是cos[(x+y) - x ]=cosy 我看不懂.所以详细点儿 -
侯雄发13236529164 ______ 这是个公式 cos(x+y)=cosxcosy-sinxsiny cos(x-y)=cosxcosy+sinxsiny 现在 sinxsin(x+y)+cosxcos(x+y)=cos[x-(x+y)]=cosy

扶炎心2331求解怎么从sinX≤y得到的:0≤X≤arc sin y或π - arc sin y≤X≤π -
侯雄发13236529164 ______[答案] sinx在[-π/2,π/2]上递增,arcsinx也是递增的,且范围是[-π/2,π/2] 对sinX≤y两边同时作用arcsin得到X≤arc sin y sinx=sin(π-x) 就可以得到π-x≤arcsiny 至于0和π两个界限应该是额外规定的,不是这个不等式解出的,否则sinx有周期,范围应该更大

扶炎心2331(sinx+siny)^2+(cosx+cosy)^2为什么等于1+1+2(sinxsiny+cosxcosy)=2+2cos(x - y) -
侯雄发13236529164 ______ 解: (sinx+siny)^2+(cosx+cosy)^2 = sinx^2 + siny^2 + 2sinxsiny + cosx^2 + cosy^2 + 2cosxcosy =(sinx^2 + cosx^2)+(siny^2 + cosy^2)+ 2sinxsiny + 2cosxcosy = 1 + 1 + 2 (sinxsiny + cosxcosy) 再根据 余弦差交公式 sinxsiny+cosxcosy = cos(x-y) 即得 (sinx+siny)^2+(cosx+cosy)^2 = 1 + 1 + 2 (sinxsiny + cosxcosy) = 2 + 2cos(x-y)

扶炎心2331求y=sinx - arc cos x和y=acr sin x + arc tan x的定义域 -
侯雄发13236529164 ______ 1]=[-1,1];值域[-3π/4,3π/y=sinx-arc cos x的定义域为R∩[-1;值域[-sin1+π,sin1];非奇非偶 定义域内单增 y=acr sin x + arc tan x的定义域为[-1,1]∩(-∞,∞)=[-1,1]

扶炎心2331化简:sin(x+y)cosx - 1/2[sin(2x+y) - siny] -
侯雄发13236529164 ______ sin(2x+y)=sin(x+y+x)=sin(x+y)*cosx+cos(x+y)*sinx siny=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

扶炎心2331设xsiny+ysinx=y 求y' siny+xcosy*y'+y'*sinx+ycosx=0 为啥cosy后面还加个y' sinx前面加个y'? -
侯雄发13236529164 ______[答案] (ysin x)'=sin x+ycos x

扶炎心2331微分方程y＂+y'=x+sinx通解,要过程 -
侯雄发13236529164 ______ 特征方程 r^2+r=0 r=0,r=-1 因此齐次通解是 y=C1x+C2e^(-x) 特解分为两部分 y＂+y'=x和y＂+y'=sinx 对于y＂+y'=x 设特解为y=x(ax+b) y'=2ax+b y''=2a 代入方程得2a+2ax+b=x a=-1/2,b=1 y=x(-1/2x+1) 下面求y＂+y'=sinx 设特解为y=acosx+bsinx y'=-...

扶炎心2331求证:sin(x+y)cos(x - y)=sinxcosx+sinycosy -
侯雄发13236529164 ______ sin(x+y)cos(x-y)=(sinxcosy+cosxsiny)(cosxcosy+sinxsiny)=sinxcosxcosy^2+sinx^2cosysiny+cosx^2sinycosy+cosxsinxsiny^2=sinxcosx(cosy^2+siny^2)+cosysiny(sinx^2+cosx^2)=sinxcosx+sinycosy