高一数学函数题目
S=cosα^2+sinβ^2 = 1 +[cos(2α) -cos(2β)]/2 = 1 - sin(α+β)sin(α-β)
= 1 - sin(2π/3)sin(2π/3-2β) = 1 - [(genhao3)/2]
α,β∈[0,π/2] ===> -π/3 -(genhao3)/2 = S >= 1 - (genhao3)/2
即,值域为:[7/4,1-(genhao3)/2]。
S=cosα^2+sinβ^2 = 1 +[cos(2α) -cos(2β)]/2 = 1 - sin(α+β)sin(α-β)
= 1 - sin(2π/3)sin(2π/3-2β) = 1 - [(genhao3)/2]
α,β∈[0,π/2] ===> -π/3 -(genhao3)/2 = S >= 1 - (genhao3)/2
即,值域为:[7/4,1-(genhao3)/2]。
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