已知cosx+siny=1/2,求siny-cosx平方的最值，要具体过程！在线等！

显示全部楼层 · 2011-1-30 14:26:39

cosx + siny = 1/2
(1)Taking derivative to the both side w.r.t x-sinx + cosy dy/dx =0
dy/dx = sinx/cosyS = (siny - cosx)^2
= (cosx + siny)^2 - 4sinycosx
= 1/4 - 4sinycosxdS/dx = -4[ -siny sinx + cosx cos y dy/dx ]
= -4[ - sinxsiny + sinxcosx ]dS/dx = 0sinx[ -siny + co

千问 · 2011-1-30 14:26:39

cosx+siny=1/2 平方得(cosx+siny)^2=1/4 展开得(cosx）^2+(siny)^2+2cosxsiny=1/4因为 (cosx）^2+(siny)^2=1，所以2cosxsiny=-3/4（siny-cosx）^2=(cosx）^2+(siny)^2-2cosxsiny=1-(-3/4)=7/4