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