jijiji~已知cosx+siny=1/2,求siny-cos^2*x的最值

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jijiji~已知cosx+siny=1/2,求siny-cos^2*x的最值
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jijiji~已知cosx+siny=1/2,求siny-cos^2*x的最值
jijiji~已知cosx+siny=1/2,求siny-cos^2*x的最值

jijiji~已知cosx+siny=1/2,求siny-cos^2*x的最值
cosx+siny=1/2
cosx=1/2-siny
siny-cos²x=siny-(1/2-siny)²
=siny-sin²y+siny-1/4
=-sin²y+2siny-1/4
=-(siny-1)²+3/4
当siny=1时,siny-cos²x有最大值3/4;当siny=-1时,siny-cos²x有最小值-13/4.