Question

已知sinx+siny=1/3,求ω=siny-cos2x的最值

Answer 1

sinx+siny=1/3
siny=1/3-sinx
由-1≤siny≤1,-1≤sinx≤1得
-1≤1/3-sinx≤1
-2/3≤sinx≤1
ω=siny-cos2x
=1/3-sinx-2(sinx)^2-1
=-2(sinx)^2-sinx-2/3
=-2(sinx+1/4)^2-13/24
sinx=-1/4时，有ωmax=-13/24
sinx=1时，有ωmin=-11/3

Answer 2

ω=siny+(sinx)^2-1
=1/3-sinx+(sinx)^2-1
=(sinx)^2-sinx-2/3
=(sinx-1/2)^2-2/3-1/4
=(sinx-1/2)^2-11/12
最大值：sinx=-1时，为5/4
最小值：sinx=1/2时，为-11/12

Answer 3

ω=siny-cos2x
=1/3-sinx-cos2x
=2sin²x-sinx-2/3
=2(sin²x-0.5sinx)-2/3
=2(sinx-1/4)²-19/24
所以ω的最小值为-19/24，最大值为56/75

Answer 4

sinx+siny=1/3
siny=1/3-sinx
由-1≤siny≤1,-1≤sinx≤1得
-1≤1/3-sinx≤1
-2/3≤sinx≤1
ωmax=4/9,wmin=-11/12
绝对准确哦！