Question

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

Answer 1

cosx + siny = 1/2 (1)
Taking derivative to the both side w.r.t x
-sinx + cosy dy/dx =0
dy/dx = sinx/cosy
S = (siny - cosx)^2
= (cosx + siny)^2 - 4sinycosx
= 1/4 - 4sinycosx
dS/dx = -4[ -siny sinx + cosx cos y dy/dx ]
= -4[ - sinxsiny + sinxcosx ]
dS/dx = 0
sinx[ -siny + cosx ] =0
cosx = siny
or sinx =0
siny = cosx
from (1)
2siny = 1/2
siny = 1/4
or sinx=0 => x =0
from (1)
1+siny = 1/2
siny = -1/2
max S at siny = -1/2 , cosx = 1
max S = 1/4 - 4(-1/2)(1)
= 3/8
minS = 1/4 - 4(1/4)(1/4)
= 3/16

Answer 2

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