已知cosx+cosy=1/3 ,求cosx-sin^2y的最大值和最小值

cosx+cosy=1/3,即cosx=1/3 -cosy,
所以
cosx-sin^2y
=cosx - (1-cos^2y)
=1/3 -cosy -1 +cos^2y
=cos^2y -cosy -2/3
=(cosy -1/2)^2 - 11/12
所以当cosy =1/2时,
cosx-sin^2y取最小值 -11/12,
而当cosy= -1时,
cosx-sin^2y取最大值 (-1-1/2)^2 -11/12= 4/3