y=sin²x sin2x 3cos²x
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y=Asin(ωx+ψ)周期为:T=2π/ωy=sin(x+1),ω=1,所以T=2π
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)=(si
y=sin(sinx)y‘=cos(sinx)*(sinx)'=cos(sinx)*cosx
设sin2x+1=t(1≤t≤2)则函数为y=t+9t-1(1≤t≤2),∵y′=1-9t2<0在1≤t≤2时恒成立,故y=t+9t-1在[1,2]为减函数,当t=1时,y取最大值9,当t=2时,y取
1.确定定义域{x|x∈R}2.判断奇偶性y(-x)=y(x)偶函数3.画图像画出x>0部分图像,将x>0部分图像向左翻折再问:那如果x属于-2π到2派间应怎样取值(5点法)再答:从右面开始画x-3π
你算错了,导数分子是1,(cosx)′=-sinx;导数应是1/sin²x;
y=sin^3x是复合函数可以设t=sinxt'=cosxy=t^3y'=3t^2*t'y'=3sin^2x*cosx
y'={[(sinx)^n][(cosx)^n]}'=ncosx(sinx)^(n-1)-nsinx(cosx)^(n-1)
∵y=x^(sinx)∴lny=sinx*lnx两边求导:y'/y=cosx*lnx+(sinx)/x∴y'=y[cosx*lnx+(sinx)/x]=x^(sinx)[cosx*lnx+(sinx)
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
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)
y'sin(y/x)-y/x*sin(y/x)+1=0令y/x=u,则y'=u+xu'所以(u+xu')sinu-usinu+1=0xu'sinu+1=0-sinudu=dx/x两边积分:cosu=l
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)/
-2k=cos2x-cos2y=[2(cosx)^2-1]-[2(cosy)^2-1]=2[(cosx)^2-(cosy)^2]cos^2x-cos^2y=-k
(1)当y=C时,sin[(x+C)/2]=sin[(x-C)/2]移项,和差化积有2cos{[(x+C)/2+(x-C)/2]/2}sin{[(x+C)/2-(x-C)/2]/2}=0,即cos(x
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