Question

急！一道高中数学题，详细解释

已知3sin^2(A)+2sin^2(B)-2sin(A)=0.则y=sin^2(A)+sin^2(B)的最大值为_______
答案4/9

Answer 1

解：3sin²A+2sin²B-2sinA=0，则3(sinA-1/3)²=1/3-2sin²B
∵sin²B∈[0,1]，则1/3-2sin²B∈[-5/3,1/3]
∵3(sinA-1/3)²≥0，则3(sinA-1/3)²∈[0,1/3]，则sinA-1/3∈[-1/3,1/3]，则sinA∈[0,2/3]
又∵3sin²A+2sin²B-2sinA=0，即2(sin²A+sin²B)=2sinA-sin²A=1-(sinA-1)²
则y=sin²A+sin²B=1/2-1/2(sinA-1)²∈[0,4/9]
则y的最大值为4/9

Answer 2

令t=sin^2(A)
sin^2(B)=t-3t^2/2
y=t^2+t-3t^2/2
=—t^2/2+t
当t=1，即sin（A）=1或-1，即A=kπ+π/2
y最大=1/2