【高一数学】急

(sin b)^2=[2cosa-(sin a)^2]/2
=[2cos a-(sin a)^2]/2
=[2cos a-1+(cos a)^2]/2
=[(cos a+1)^2-2]/2.
因为0<=(sin b)^2<=1,
所以0<=(cos a+1)^2-2<=2,
即2<=(cos a+1)^2<=4,
所以-2<=cos a+1<=-根号2
或根号2<=cos a+1<=2,
因为-1<=cos a<=1,
所以只有根号2<=cos a+1<=2,即根号2-1<=cos a<=1
y=(sin a)^2+(sin b)^2
=[2cosa-(sin a)^2]/2+(sin a)^2
=[2cosa+1-(cos a)^2]/2
=-1/2(cosa-1)^2+1
M=2√2-2,N=1
M+N=2√2-1

是-2么?
消元啊,由sin²α+2sin²β=2cosα
得sin²β=cosα-sin²α/2∈[0,1] (消元时注意控制范围)
由上式右端解得cosa∈[-3,1] (虚惊一场)
带入到y=sin²α+sin²β,得
y=-1/2*(cosa-1)² cosa∈[-1,1]
所以y∈[-2,0]
所以......
(但愿我没算错数)

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