Question

已知sinx+siny=1/3 ,求z=sinx-cos^2y得取值范围

Answer 1

已知sinx+siny=1/3

则 z=sinx-cos^2y=1/3-siny-cos2y

=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