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+α^{2}+β^{2}+γ^{2}
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
Δ^{2} = 1α^{2}β^{2}γ^{2}+2αβγ.
Also, sin(D/2) = Δ/4cosh(½a)cosh(½b)cosh(½c)
proof
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) = βγα
sinh(a)sinh(c)cos(B) = cosh(a)cosh(c)  cosh(b) = αγβ,
sinh(b)sinh(a)cos(C) = cosh(b)cosh(a)  cosh(c) = βαγ.
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 Δ^{2} = 1α^{2}β^{2}γ^{2}+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
(sinh(a)sinh(b)sinh(c))^{2}cos(A+B+C)
= (βγα)(αγβ)(βαγ)  Δ^{2}(βγα+αγβ+βαγ)
= (α1)(β1)(γ1)
(α+β+γ+αβ+βγ+γα+α^{2}+β^{2}+γ^{2}αβγ),
(after much simplification  I used Maple to help).
Now sinh^{2}(a) = cosh(a)^{2}1 = α^{2}1, so we get
cos(A+B+C) =
(α+β+γ+αβ+βγ+γα+α^{2}+β^{2}+γ^{2}αβγ)/
(1+α)(1+β)(1+γ).
By the gaussbonnet 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 :
tan^{2}(x/2) = (1cos(x))/(1+cos(x)),
If we use this to derive tan^{2}(D/2) from cos(D) and simplify considerably, we get
tan^{2}(D/2) = Δ^{2}/(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 = 2sin^{2}(D/2), so that, after tidying,
2sin^{2}(D/2) = Δ^{2}/(1+α)(1+β)(1+γ). As 1+α = 2 cosh^{2}(½a), etc, we get the result.
