hyperbolic area - proof of heron's formula

heron's formula for hyperbolic triangles
If a hyperbolic triangle has sides of lengths a, b and c, then its area D is given by
cos(D) = (α+β+γ+αβ+βγ+γα+α222- αβγ)/(1+α)(1+β)(1+γ), and hence by
tan(D/2) = Δ/(1+α+β+γ),
where α=cosh(a),β=cosh(b),γ=cosh(c), and Δ2 = 1-α222+2αβγ.
Also, sin(D/2) = Δ/4cosh(½a)cosh(½b)cosh(½c)

Standard notation
If ABC is a hyperbolic triangle, then we write
A for the angle at vertex A, i.e. <BAC, and
a for the hyperbolic length of the side opposite A, i.e. d(B,C).

The cosine rule for hyperbolic triangles gives
sinh(b)sinh(c)cos(A) = cosh(b)cosh(c) - cosh(a) = βγ-α
sinh(a)sinh(c)cos(B) = cosh(a)cosh(c) - cosh(b) = αγ-β,
sinh(b)sinh(a)cos(C) = cosh(b)cosh(a) - cosh(c) = βα-γ.

The sine rule for hyperbolic triangles gives
sinh(b)sinh(c)sin(A) = Δ,
sinh(a)sinh(b)sin(C) = Δ,
sinh(a)sinh(c)sin(B) = Δ,
where Δ is the positive root of Δ2 = 1-α222+2αβγ.

Standard trigonometric addition formula give
cos(A+B+C) = cos(A)cos(B)cos(C)-cos(A)sin(B)sin(C) -sin(A)cos(B)sin(C)-sin(A)sin(B)cos(C).

If we multiply the last by (sinh(a)sinh(b)sinh(c))2, and substitute for the sines and cosines,
we get
= (βγ-α)(αγ-β)(βα-γ) - Δ2(βγ-α+αγ-β+βα-γ)
= -(α-1)(β-1)(γ-1) (α+β+γ+αβ+βγ+γα+α222-αβγ),
(after much simplification - I used Maple to help).

Now sinh2(a) = cosh(a)2-1 = α2-1, so we get
cos(A+B+C) = -(α+β+γ+αβ+βγ+γα+α222-αβγ)/ (1+α)(1+β)(1+γ).

By the gauss-bonnet formula, the area D is given by π-(A+B+C),
so cos(D) = - cos(A+B+C), which gives the first result.

The second is derived by using the identity :

tan2(x/2) = (1-cos(x))/(1+cos(x)),

If we use this to derive tan2(D/2) from cos(D) and simplify considerably, we get
tan2(D/2) = Δ2/(1+α+β+γ)2. As D/2 < π/2, and Δ > 0, we can then take positive roots.

The final result follows from the fact that cos(D)+1 = 2sin2(D/2), so that, after tidying,
2sin2(D/2) = Δ2/(1+α)(1+β)(1+γ). As 1+α = 2 cosh2(½a), etc, we get the result.

heron's formula