area of a right-angled hyperbolic triangle
If the hyperbolic triangle ABC has a right angle at A, and
d(A,B)=c, d(B,C)=a, d(C,A)=b, then its hyperbolic area D
is given by
sin(D) = sinh(b)sinh(c)/(cosh(a)+1).
proof
By Pythagoras's Theorem, cosh(a) = cosh(b)cosh(c).
Let <ABC=B, <ACB =C.
By the gauss-bonnet formula, D = π/2-(B+C).
Since this is in the range (0,π/2),
it is determined by the
value of its sine
or cosine.
Also,
cos(B) = tanh(c)/tanh(a) = sinh(c)cosh(a)/sinh(a)cosh(c)
= sinh(c)cosh(b)/sinh(a).
sin(B) = sinh(b)/sinh(a).
Similarly,
cos(C) = sinh(b)cosh(c)/sinh(a), sin(C) =sinh(c)/sinh(a).
By the addition formula,
cos(B+C)
= (sinh(c)cosh(a)sinh(b)cosh(b)-sinh(b)sinh(c))/sinh2(a)
= sinh(b)sinh(c)(cosh(a)+1)/(cosh2(a)-1)
= sinh(b)sinh(c)/(cosh(a)+1).
As D = π/2-(B+C), sin(D) = cos(B+C), and we are done.
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