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已知sinx+siny=1/3,求M=sinx-cos²y的最大值和最小值.

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已知sinx+siny=1/3,求M=sinx-cos²y的最大值和最小值.
t=sinx.
siny = 1/3 - sinx = 1/3 - t,
M = sinx - [cosy]^2 = t - 1 + [siny]^2 = t - 1 + [1/3 - t]^2 = t - 1/3 - 2/3 + (t-1/3)^2
= (t-1/3)^2 + (t-1/3) + 1/4 - 1/4 - 2/3
= [t-1/3 + 1/2]^2 - 11/12
= [t + 1/6]^2 - 11/12,
t=sinx=-1/6时,M取得最小值-11/12,
t=sinx=1时,M取得最大值[1+1/6]^2 - 11/12 = 49/36 - 11/12 = 16/36 = 4/9