y=sin^2x 2sinxcosx-3cos^2x

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y=sin^2x 2sinxcosx-3cos^2x
三角函数最值问题已知x,y,z为实数,求:f(x,y,z)=[sin(x-y)]^2+[sin(y-z)]^2+[sin

sin^2(x-y)+sin^2(y-z)+sin^2(z-x)=[1-cos2(x-y)+1-cos2(y-z)+1-cos2(z-x)]/2=3/2-[(cos2xcos2y+sin2xsin2y

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)^2-(siny)^2.

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=COS^2角*SIN角当sin角=多少y有最大值

y=cos²xsinxy²=cos²xcos²xsin²x=1/2(cos²xcos²x*2sin²x)≤1/2*[(2

tan(x+y)tan(x-y)=sin^2x-sin^2y/cos^2x-sin^2y 顺便问一下. tan,sin,

tan,正切;sin,正弦;cos,余弦tan(x+y)tan(x-y)=sin(x+y)/cos(x+y)*sin(x-y)/cos(x-y)=sin(x+y)sin(x-y)/[cos(x+y)c

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

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 =(cos^2) x - sin (3^x),求y'

y'=(cos²x)'-(sin3^x)'=2cosx·(cosx)'-cos3^x·(3^x)'=2cosx·(-sinx)-cos3^x·(3^x·ln3)=-sin2x-ln3·cos

sin^2x+cos^2y=1/2 求3sin^2x+sin^2y的最值

sin^2x+cos^2y=1/2∴sin^2x=1/2-cos^2y3sin^2x+sin^2y=3(1/2-cos^2y)+sin^2y=1.5-3cos^2y)+sin^2y又有sin^2y+c

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

求微分方程y'+sin[(x+y)/2]=sin[(x-y)/2]通解

(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

求导:x^2*y^2 + x sin(y) = 1

对这样的隐函数求导数的时候,就把y看作x的函数,y对x求导就得到dy/dx所以原等式对x求导得到2xy²+x²*2y*dy/dx+siny+x*cosy*dy/dx=0于是化简得到

求证cosx-cosy=-2sin (x+y/2)*sin (x-y/2)

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

y=sin(2x+30度)求导,y'=?

过程:先将括号里的当作一个整体,即求sinx的导数,所以是cos(2x+30度),再对括号里的求导,所以得2由复合函数的求导法则,知y=2cos(2x+30度)

函数y=sin(π2

y=sin(π2+x)cos(π6-x)=cosx(32cosx+12snx)=32cos2x+12sinxcosx=34(1+cos2x)+14sin2x=12sin(2x+π3)+34∴T=2π2

求证1、sin l-sin y=2cos(l+y)/2 sin(l-y)/2

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

求y=sin(e^2x)的微分y'

y'=2e^2xcos(e^2x)把y看成复合函数sint,t=e^m,m=2x.复合函数求导,等于三个分别求导的积

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

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