求方程最值

T=siny-(cosx)^2
=1/3-sinx-1+(sinx)^2
=(sinx-1/2)^2-1/4+1/3-1
=(sinx-1/2)^2-11/12
因为,sinx+siny=1/3,sinx=1/3-siny
-1<=siny<=1,-1<=-siny<=1
-2/3<=1/3-siny<=4/3
所以-2/3<=sinx<=1
则,-7/6<=sinx-1/2<=1/2
0<=(sinx-1/2)^2<=49/36
-11/12<=(sinx-1/2)^2-11/12<=49/36-11/12=4/9
即-11/12<=T=4/9
故T=siny-(cosx)^2的最大值4/9,最小值-11/12