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已知sinx+siny=1/3 ,求z=sinx-cos^2y得取值范围
请写出具体过程!
人气:486 ℃ 时间:2019-12-10 00:59:06
解答
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
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