若sin2α+2sin2β=2cosα,y=sin2α+cos2β的最大值为M,最小值为m.求M+m.

y = (sinα)^2 + (cosβ)^2
= (sinα)^2 + [1 - (sinβ)^2] (利用sin^2+cos^2=1)
= (sinα)^2 + [1 - (cosα - 1/2*(sinα)^2)] (利用条件)
= 1 - (cosα)^2 + 1 - cosα + 1/2 - 1/2*(cosα)^2 (利用sin^2+cos^2=1)
= 5/2 - 3/2*(cosα + 1/3)^2 + 1/6 (配方)
根据条件,我们知道2cosα - (sinα)^2 = 2(sinβ)^2 ≥ 0,且 cosα ≥ 0,化简之后,我们有1 ≥ cosα ≥ √2 - 1,所以,我们可以得到,
m = 0,此时,cosα = 1,
M = 2√2 - 1,此时,cosα = √2 - 1.
所以,M + m = 2√2 - 1