the function Δ
For complex a,b,c, we define
Δ^{2} = Δ^{2} (a,b,c)= 1 + 2 cosh(a)cosh(b)cosh(c)  cosh^{2}(a)  cosh^{2}(b) 
cosh^{2}(c), and
s = s(a,b,c) = ½(a+b+c).
(Δ1) Δ^{2} = 4sinh(s)sinh(sa)sinh(sb)sinh(sc),
(Δ1') Δ^{2} = sinh^{2}(a)sinh^{2}(b)(cosh(a)cosh(b)cosh(c))^{2}, etc.
the function Φ
For complex α,β,γ
Φ^{2} = Φ(α,β,γ) =
1 + 2cos(α)cos(β)cos(γ)+cos^{2}(α)+cos^{2}(β)+cos^{2}(γ), and
σ = σ(α,β,γ) = ½(α+β+γ).
(Φ1) Φ^{2} = 4cos(σ)cos(σα)cos(σβ)cos(σγ),
(Φ1') Φ^{2} = sin^{2}(α)sin^{2}(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))^{2}Δ^{2}= Φ^{4}.
(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))^{2}Φ^{2}= Δ^{4}.
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) Δ^{2} > 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) Φ^{2} > 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 = Φ^{2}/sin(α)sin(β)sin(γ),
(R2) Φ = Δ^{2}/sinh(a)sinh(b)sinh(c),
(R3) ΔΦ = sin(α)sin(β)sin(γ)sinh(a)sinh(b)sinh(c),
(R4) Δ^{3} = sin(α)sin(β)sin(γ)(sinh(a)sinh(b)sinh(c))^{2},
(R5) Φ^{3} = (sin(α)sin(β)sin(γ))^{2}sinh(a)sinh(b)sinh(c).
proofs
(Δ2) Since a,b,c are the lengths of the sides of a triangle, s,sa,sb,sc 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,σα,σβ,σγ 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.
(R15)
If we use the first two parts of (Φ3) and multiply, we get Φ^{2} = sin(β)sinh(a)sinh(c)sin(α)sin(β)sin(γ).
Hence, using (Δ3), Φ^{2} = 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).
