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三角函数问题

　cos²α + cos²β
= (1+cos2α)/2 + (1+cos2β)/2
= (cos2α+cos2β)/2 + 1
= cos(α+β)cos(α-β) + 1
= cos(2π/3)cos[α - (2π/3 - α)] + 1
= -(1/2)cos(2α - 2π/3) + 1

由 α + β = 2π/3 得 β = 2π/3 - α
因为 α、β 都是锐角，
所以 0 < α < π/2 ， 且 0 < 2π/3 - α < π/2
得 π/6 < α < π/2
从而 -π/3 < 2α - 2π/3 < π/3
故　cos(2α - 2π/3) ∈ (1/2, 1]
故　-(1/2)cos(2α - 2π/3) ∈ [-1/2, -1/4)
故　-(1/2)cos(2α - 2π/3) + 1 ∈ [1/2, 3/4)

即　cos²α + cos²β 的取值范围是 [1/2, 3/4)