second existence theorem
If α,β,γ > 0, there is a hyperbolic triangle with angles α,β,γ if and only if α+β+γ<π.
proof
The only if part is trivial.
Now suppose that 2σ = α+β+γ<π with α,β,γ > 0.
Let a,b,c be defined by cosh(a) = (cos(β)cos(γ)+cos(α))/sin(α)sin(β), etc.
From (Φ1), Φ2(α,β,γ) > 0 since σ,σ-α,σ-β,σ-γ
are in (-½π,½π).
From (ΔΦ), Δ2(a,b,c) > 0. Now, by the first existence theorem, there is a hyperbolic
triangle with sides of length a,b,c. The angles of this triangle are α',β',γ', where
cos(α') = (cosh(b)cosh(c)-cosh(a)/sinh(b)sinh(c), by the First Cosine Rule for the triangle.
If we now substitute for cosh(a), cosh(b), cosh(c) in terms of α,β,γ, we get (eventually)
cos(α') = cos(α)Φ2/sinh(b)sinh(c)sin(α)2sin(β)sin(γ).
Also, from sinh2(x) = cosh2(x)-1, we get, on taking positive roots
sinh(b) = Φsin(α)/sin(β), sinh(c) = Φ/sin(α)sin(β).
Combining these three results, cos(α')=cos(α), so α' = α as each is in (0,π).
Similarly, β' = β, γ' = γ, so we have a triangle of the required type.
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