若(sin a)^2+2(sin b)^2=2cos a,y=(sin a)^2+(sin b)^2的最大值为M,最小值为m,则M+m=?答案说是2倍根号2减1，怎么算？

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若(sina)^2+2(sinb)^2=2cosa,y=(sina)^2+(sinb)^2的最大值为M,最小值为m,则M+m=?答案说是2倍根号2减1，怎么算？若(sina)^2+2(sinb)^2=

若(sin a)^2+2(sin b)^2=2cos a,y=(sin a)^2+(sin b)^2的最大值为M,最小值为m,则M+m=?答案说是2倍根号2减1，怎么算？
若(sin a)^2+2(sin b)^2=2cos a,y=(sin a)^2+(sin b)^2的最大值为M,最小值为m,则M+m=?
答案说是2倍根号2减1，怎么算？

若(sin a)^2+2(sin b)^2=2cos a,y=(sin a)^2+(sin b)^2的最大值为M,最小值为m,则M+m=?答案说是2倍根号2减1，怎么算？
(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


(1)得sinb^2=cosa-1/2*(sina^2)
(2)代入y=1/2*sina^2+cosa
=1/2(1-cosa^2)+cosa
=-1/2*(cosa-1)^2+1
∵cosa∈[-1,1]
cosa=1 ymax=1=M,
cosa=-1 ymin=-1=m
∴M+m=0

因为(sin a)^2+2(sin b)^2=2cos a，
所以(sin b)^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,

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因为(sin a)^2+2(sin b)^2=2cos a，
所以(sin b)^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,
所以y=(sin a)^2+(sin b)^2
=1-(cos a)^2+[2cos a-1+(cos a)^2]/2
=-(cos a)^2/2+cos a+1/2
=-(cos a-1)^2/2+1,
因为根号2-1<=cos a<=1，
所以根号2-2<=cos a-1<=0,
所以0<=(cos a-1)^2<=(根号2-2)^2=6-4根号2，
所以2根号2-3<=-(cos a-1)^2/2<=0,
所以2根号2-2<=y<=1,
所以M+m=2根号2-2+1=2根号2-1.

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