Y=sin(5-3X)的单增区间-搜答案-拍题作业网

sin(3x+5π/2)=sin(3x+π/2)=cos3xcosx对称轴就是sinx取最值的地方即x=kπ则此处是3x=kπx=kπ/3

y=sin(3x+π/12)sin(3x-5π/12)=sin(π/2-3x-π/12)sin(3x-5π/12)=cos(5π/12-3x)sin(3x-5π/12)=cos(3x-5π/12)si

求函数y=sin²x+3sinxcosx+5cos²x的值域y=1-cos²+3sinxcosx+5cos²x=4cos²x+3sinxcosx+1=

sinx的周期是2pai,sin3x的周期是三分之二pai,sin5x的周期是五分之二pai取其最小公倍数,则y的周期是2pai.

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

你用积化和差公式一套,然后就能看出它的最小正周期来的.应该是1pi

y=3sin（x+20°）+5sin（x+80°）=3sin（x+20°）+5sin（x+20°+60°）=3sin（x+20°）+5[sin（x+20°）cos60°+cos（x+20°）sin60

y=sin(3x+5)的导数

y'=sin‘(3x+5)×(3x)'=3cos(3x+5)

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=sin^3x+sinx^3y'=3sin²xcosx+cosx³*3x²=3sin²xcosx+3x²cosx³

复合函数应该用链式法则求导：若y=g(u),u=f(x),则dy/dx=dy/du*du/dxy=sin^5xdy/dx=dsin^5x/dsinx*dsinx/dx=5sin⁴xcosx

再答：希望采纳！谢谢

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

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

lny=sinxlnx对x求导(1/y)*y'=cosx*lnx+sinx*(1/x)y'=y*[cosx*lnx+sinx*(1/x)]所以y'=x^sinx*[cosx*lnx+sinx*(1/x

y=sin(x+π/3)sin(x+π/2)=sin(x+π/3)cosx=(sinxcosπ/3+cosxsinπ/3)cosx=1/2sinxcosx+√3/2cos^2(x)[cos^2(x)指

y=(lnx)^3+(sinx)^2y'=dy/dx=3(lnx)^2/x+2sinxcosx=3(lnx)^2/x+sin2xdy=[3(lnx)^2/x+sin2x]dx

y=3sin(x+20°)+5sin(x+80°）=3sin(x+20°)+5sin(x+20°+60°）=3sin(x+20°)+5*[sin(x+20°)*cos60°+cos(x+20°)*si

从周期的定义来看,令f(x)=sin3x+cos5x,f(x+2Pi)=sin(3x+6Pi)+cos(5x+10Pi)=sin3x+cos5x=f(x)所以周期为2Pi,其实只要找到一个T即是sin

π+kπ