设cos(α-β)=-4/5,sin（α+β）=-5/13,α∈（π,3/2π）,β∈（2/π,π）,求cos2α,co

α-β∈(0,π)
cos(α-β) = -4/5,所以 sin(α-β) = 3/5
α+β∈(3π/2,5π/2)
sin(α+β) = -5/13,所以 cos(α+β) = 12/13
和差化积公式：
sin(α+β) + sin(α-β) = 2 sinα cosβ = -5/13 + 3/5 = 14/65
cos(α+β) + cos(α-β) = 2 cosα cosβ = 12/13 - 4/5 = 8/65
两式相除：tanα = 7/4
所以：cosα = -4/sqrt(65),其中sqrt()代表开根号
代回得：cosβ = (8/65) / (2 cosα) = -1/sqrt(65)
cos2α = 2(cosα)^2 - 1 = 32/65 - 1 = -33/65
cos2β = 2(cosβ)^2 - 1 = 2/65 - 1 = -63/65