已知sinx+siny=1/3求sinx-cos^2y的最大值和最小值

问题描述:

已知sinx+siny=1/3求sinx-cos^2y的最大值和最小值

sinx+siny=1/3
sinx=1/3 -siny
-1≤sinx≤1
-1≤1/3-siny≤1
-2/3≤siny≤4/3
又-1≤siny≤1,因此-2/3≤siny≤1

sinx-cos²y=1/3-siny-(1-sin²y)=sin²y-siny -2/3=(siny -1/2)² -11/12
当siny=-2/3时,sinx-cos²y有最大值(sinx-cos²y)max=(-2/3 -1/2)²-11/2=4/9
当siny=1/2时,sinx-cos²y有最小值(sinx-cos²y)min=-11/2

由sinx+siny=1/3,得sinx=1/3-siny
原式z=1/3-siny-cos^2y
=1/3-siny-1+2sin^2y
=2[sin^2y-0.5siny-(1/3)]
=2(siny-1/4)^2-1/8-2/3
=2(siny-1/4)^2-19/24
因为-1≤siny≤1,则 -1-1/4=-5/4≤siny-1/4≤3/4
9/8≤2(siny-1/4)^2≤25/8
1/4≤2(siny-1/4)^2-19/24≤7/3
即,1/4≤z≤7/3

cos^2y=1-sin^2(y)
sinx=1/3-siny
原式=1/3-siny-(1-sin^2y)
令siny=t∈[-1,1]
则f(t)=t^2-t-2/3,t∈[-1,1]
二元一次方程,定义域在-1,1
min在x=1/2取到,为-11/12
max在x=-1取到,为4/3
∴原式最大值为4/3,最小值-11/12