y=sin(x0 △X)-sinx0=2sin△x 2cos(x0 △ 2)

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证明数列极限|cosx-cosx0|=|-2sin(x+x0)/2*sin(x-x0)/2|

和差化积公式.如果不清楚,请去翻翻书.

求y=sin(x+1)周期

y=Asin(ωx+ψ)周期为:T=2π/ωy=sin(x+1),ω=1,所以T=2π

matlab画y=sin(x)+sin(2*x)+...+sin(20*x)的图像

x=0:0.01:1;y=0;fori=1:20y=y+sin(i*x);endplot(y);

证明sin(x+y)sin(x-y)=sinx-siny

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|x|取值画图

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

y=sin^3x的导数

y=sin^3x是复合函数可以设t=sinxt'=cosxy=t^3y'=3t^2*t'y'=3sin^2x*cosx

y=x^sin的导数

∵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)

已知函数f(x)=2sinx*sin(π/2+x)-2sin^2x+1 .(2)若f(X0/2)=√2/3,X0∈(-π

f(x)=2sinx*sin(π/2+x)-2sin^2x+1=2sinxcosx+cos2x=sin2x+cos2x=√2sin(2x+π/4)因为f(x0/2)=根2/3所以sin(x0+π/4)

一道三角恒等式证明题请证明sin(x+y)sin(x-y)=sin^2(x)-sin^2(y)

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

sin(x)^y求导

多么庞大,在牧场!在我们之间乌鸫飞翔后代的眼睛这么快就开始我将利用水.今天空气中什么都没有不是他们你得起的誓言哈哈

y=sin[sin(x^2)] 则dy/dx=?

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-1 x 求导

dx/dt={1/[2√(1-t^2)]}(-2t)=-t/√(1-t^2)dy/dt=1/√(1-t^2)dy/dx=[1/√(1-t^2)]/[-t/√(1-t^2)]=-1/t再问:为啥dy/d

x*y'*sin(y/x)-y*sin(y/x)+x=0 求微分方程的解

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

已知函数f(x0=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)

/>(1)f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]=cos(2x-π/3)+2sin

求导 y=sin(x^x)

(x^x)'=(1+lnx)x^x所以y'=(1+lnx)x^xcos(x^x)

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

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)sin(x-y)=k,求cos^2x-cos^2y

-2k=cos2x-cos2y=[2(cosx)^2-1]-[2(cosy)^2-1]=2[(cosx)^2-(cosy)^2]cos^2x-cos^2y=-k

sin(x.y)中,

点的坐标表示经过点(x,y)的正切值再答:正弦值再答:刚打错了不好意思再问:一语点醒梦中人,谢谢再答:呵呵

.把函数y=cos(3x+π\4) 的图象适当变换可以得到y=sin(-3x0 的图象,这种变换可以是

y=sin(-3x)=-sin3xy=cos(3x+π\4)=cos(π/2+3x-π/4)=-sin(3x-π/4)=-sin[3(x-π/12)]所以,要向左平移π/12个单位再问:不是应该左加右