proofs from further trigonometry for hyperbolic geometry

the function Δ
For complex a,b,c, we define
Δ2 = Δ2 (a,b,c)= 1 + 2 cosh(a)cosh(b)cosh(c) - cosh2(a) - cosh2(b) - cosh2(c), and
s = s(a,b,c) = ½(a+b+c).
(Δ1) Δ2 = 4sinh(s)sinh(s-a)sinh(s-b)sinh(s-c),
(Δ1') Δ2 = sinh2(a)sinh2(b)-(cosh(a)cosh(b)-cosh(c))2, etc.
the function Φ
For complex α,β,γ
Φ2 = Φ(α,β,γ) = -1 + 2cos(α)cos(β)cos(γ)+cos2(α)+cos2(β)+cos2(γ), and
σ = σ(α,β,γ) = ½(α+β+γ).

(Φ1) Φ2 = 4cos(σ)cos(σ-α)cos(σ-β)cos(σ-γ),
(Φ1') Φ2 = sin2(α)sin2(β)-(cos(α)cos(β)+cos(γ))2, etc.

some Δ, Φ relations
(ΔΦ) If sin(α)sin(β)sin(γ) ≠ 0, let cosh(a) = (cos(β)cos(γ)+cos(α))/sin(α)sin(β), etc
Then (sin(α)sin(β)sin(γ))2Δ2= Φ4.
(ΦΔ) If sinh(a)sinh(b)sinh(c) ≠ 0, let cos(α) = (cosh(b)cosh(c)-cosh(a))/sinh(a)sinh(b), etc
Then (sinh(a)sinh(b)sinh(c))2Φ2= Δ4.

These are left as exercises on hyperbolic and euclidean trigomometry.

further hyperbolic triangle formulae
Consider the triangle ABC with a = d(B,C), b = d(C,A), c = d(A,B), α = <CAB, β = <ABC, γ = <BCA,
(Δ2) Δ2 > 0, so we may choose Δ > 0,
(Δ3) Δ = sinh(a)sinh(b)sin(γ) = sinh(b)sinh(c)sin(α) = sinh(c)sinh(a)sin(β).
(Φ2) Φ2 > 0, so we may choose Φ > 0,
(Φ3) Φ = sin(α)sin(β)sinh(c) = sin(β)sin(γ)sinh(a) = sin(γ)sin(α)sinh(b),
(R1) Δ = Φ2/sin(α)sin(β)sin(γ),
(R2) Φ = Δ2/sinh(a)sinh(b)sinh(c),
(R3) ΔΦ = sin(α)sin(β)sin(γ)sinh(a)sinh(b)sinh(c),
(R4) Δ3 = sin(α)sin(β)sin(γ)(sinh(a)sinh(b)sinh(c))2,
(R5) Φ3 = (sin(α)sin(β)sin(γ))2sinh(a)sinh(b)sinh(c).

(Δ2) Since a,b,c are the lengths of the sides of a triangle, s,s-a,s-b,s-c are positive. The result now follows from (Δ1).
(Δ3) follows from the First Cosine Rule and (Δ1') by taking positive roots.
(Φ2) Since α,β,γ are the angles of a triangle, σ,σ-α,σ-β,σ-γ ε(-½π,½π) and the result follows from (Φ1).
(Φ3) follows from the Second Cosine Rule and (Φ1') by taking positive roots.

Note that, as we have a hyperbolic triangle, a,b,c and α,β,γ are related as in (ΔΦ) and (ΦΔ) by the Cosine Rules.
If we use the first two parts of (Φ3) and multiply, we get Φ2 = sin(β)sinh(a)sinh(c)sin(α)sin(β)sin(γ).
Hence, using (Δ3), Φ2 = Δsin(α)sin(β)sin(γ), and (R1) follows. (R2) is similar from (Δ3) then (Φ3).
(R3) is obtained from the first parts of (Δ3) and (Φ3). (R4),(R5) are obtained from (R1) and (R2).

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