I was puzzeling with Area of hyperbolic triangle definition and could not figure it out, but then i thought there should be a (maybe solvable) simpler problem so here it is:


  • an hyperbolic plane with a Gaussian curvature of -1
  • on this surface there is a triangle $\triangle ABC$
  • $\angle C$ is a right angle

Then given the lengths of sides a and b what is the area?

Off course we could do:

the area $ = \frac{\pi}{2} - \angle A -\angle B$ or

$ \frac{\pi}{2} - arctan (\frac{\tanh(a)}{\sinh(b)} )- arctan (\frac{\tanh(b)}{\sinh(a)}) $

(free after http://en.wikipedia.org/wiki/Hyperbolic_triangle )

But is there not a nicer formula that does not contain any or only one trigonometric function?

(I guess the hyperbolic functions are needed)

UPDATE 07/10/2015

following https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Arctangent_addition_formula

$$ \arctan u + \arctan v = \arctan \left( \frac{u+v}{1-uv} \right) $$


$$ \frac{\pi}{2} -arctan (\frac{u}{v}) = arctan (\frac{v}{u})$$

(complement tangent = cotangent)

I get to

$$ Area = arctan(\frac{ \sinh(a)\sinh(b) - \tanh(a)tanh(b)}{\sinh(a)\tanh(a)+ \sinh(b)\tanh(b)} )$$

Can this be simplified any further? (Do i overlook the obvious again , or even did i make a mistake on the way somewhere?)

  • 1
    $\begingroup$ Is $\tan(S/2)=\tanh(a/2)\tanh(b/2)$ nice enough? $\endgroup$
    – Grigory M
    Commented Oct 3, 2015 at 19:15
  • $\begingroup$ You mean Area $ = 2 arctan ( \tanh ( a / 2 ) \tanh ( b / 2) ) $ ? It is a nice formula , can you show me the deriviation of it from where I started? (I did not manage that at all, i like to know and I would gladly accept it as an answer) $\endgroup$
    – Willemien
    Commented Oct 4, 2015 at 7:35

2 Answers 2


Let's consider an arbitrary hyperbolic triangle with angles $\alpha$, $\beta$, $\gamma$ opposite respective sides $a$, $b$, $c$. The area ($T$) is given by $$T = \pi - \alpha - \beta - \gamma \tag{0}$$ so that, writing $X_2$ for $X/2$, $$\begin{align} \cos T_2 &= \cos\left(\frac{\pi}{2}-\alpha_2-\beta_2-\gamma_2\right) = \sin\left(\alpha_2+\beta_2+\gamma_2\right) \tag{1a}\\ &= \sin\alpha_2 \cos\beta_2\cos\gamma_2+\cos\alpha_2\sin\beta_2\cos\gamma_2 \tag{1b} \\[4pt] &+\cos\alpha_2\cos\beta_2\sin\gamma_2-\sin\alpha_2\sin\beta_2\sin\gamma_2 \end{align}$$

The hyperbolic Law of Cosines tells us, for instance, $$\cos\gamma = \frac{\cosh a \cosh b - \cosh c}{\sinh a \sinh b} \tag{2}$$ so that $$\begin{align} \cos^2\gamma_2 &= \frac{1+\cos\gamma}{2} = \frac{\cosh(a+b)-\cosh c}{2\sinh a\sinh b} = \frac{\sinh s \sinh s_c}{\sinh a\sinh b} \tag{3a} \\[4pt] \sin^2\gamma_2 &= \frac{1-\cos\gamma}{2} = \frac{\cosh c-\cosh(a-b)}{2\sinh a\sinh b} = \frac{\sinh s_a \sinh s_b}{\sinh a\sinh b} \tag{3b} \end{align}$$

Likewise, $$\cos^2\alpha_2 = \frac{\sinh s \sinh s_a}{\sinh b\sinh c} \qquad \sin^2\alpha_2 = \frac{\sinh s_b \sinh s_c}{\sinh b\sinh c} \tag{4}$$ $$\cos^2\beta_2 = \frac{\sinh s \sinh s_b}{\sinh c\sinh a} \qquad \sin^2\beta_2 = \frac{\sinh s_c \sinh s_a}{\sinh c\sinh a} \tag{5}$$ From these, we have $$\begin{align} \cos T_2 &= \frac{s_0 s_b s_c+s_0 s_c s_a+s_0 s_a s_b - s_a s_b s_c}{\sinh a\sinh b\sinh c} \tag{6a}\\[4pt] &= \frac{2\sinh a_2 \sinh b_2\sinh c_2\;(1+\cosh a + \cosh b + \cosh c)}{8\;\sinh a_2 \cosh a_2\;\sinh b_2 \cosh b_2\;\sinh c_2 \cosh c_2} \end{align} \tag{6b}$$

where the numerator of $(6b)$ is a bit of a slog. (See Addendum below.) Ultimately, we conclude

$$\begin{align} \cos T_2 &= \frac{1+\cosh a + \cosh b + \cosh c}{4\;\cosh a_2\cosh b_2\cosh c_2} \tag{7a}\\[4pt] \sin T_2 &= \frac{\sqrt{1-\cosh^2 a-\cosh^2 b - \cosh^2 c+ 2 \cosh a\cosh b\cosh c}}{4\cosh a_2 \cosh b_2 \cosh c_2} \tag{7b} \end{align}$$

In the particular case of a right triangle $\gamma = \pi/2$, we have $\cosh c = \cosh a \cosh b$, and the above formulas reduce thusly:

$$\begin{align} \cos T_2 &= \frac{(1+\cosh a)(1 + \cosh b)}{4\;\cosh a_2\cosh b_2\cosh c_2} = \frac{4\cosh^2 a_2 \cosh^2 b_2}{4\cosh a_2 \cosh b_2 \cosh c_2} = \frac{\cosh a_2 \cosh b_2}{\cosh c_2} \tag{8a}\\[4pt] \sin T_2 &= \frac{\sqrt{(\cosh^2 a-1)(\cosh^2 b-1)}}{4\cosh a_2 \cosh b_2 \cosh c_2} = \frac{\sinh a\sinh b}{4\cosh a_2 \cosh b_2 \cosh c_2} = \frac{\sinh a_2 \sinh b_2}{\cosh c_2} \tag{8b} \end{align}$$

so that

$$\tan T_2 = \tanh a_2 \tanh b_2 \tag{9}$$

Addendum. Here's a walkthrough of deriving the numerator of $(5)$. I did this mostly on-the-fly; the reader is encouraged to seek a much quicker path. (Expanding everything in terms of exponentials is tedious, but effective. Of course, a computer algebra system can verify the relation in an instant.)

$$\begin{align} &\phantom{+\;} \sinh s \sinh s_b \sinh s_c + \sinh s \sinh s_c \sinh s_a \tag{Aa}\\ &+ \sinh s \sinh s_a \sinh s_b - \sinh s_a \sinh s_b \sinh s_c \\[6pt] = &\phantom{+\;} \sinh s \sinh s_c \left( \sinh s_b + \sinh s_a \right) \tag{Ab}\\ &+ \sinh s_a \sinh s_b \left( \sinh s - \sinh s_c \right) \\[6pt] = &\phantom{+\;\;} \frac12 \left(\cosh(s\;+s_c)-\cosh(s\;-s_c)\right)\cdot 2 \sinh\frac{s_a+s_b}{2}\;\cosh\frac{s_a-s_b}{2} \tag{Ac} \\[4pt] &+ \frac12 \left(\cosh(s_a+s_b)-\cosh(s_a-s_b)\right) \cdot 2\cosh\frac{s\;+s_c}{2}\;\sinh\frac{s\;-s_c}{2} \\[6pt] = &\phantom{+\;} \left(\cosh(a+b)-\cosh c\right)\cdot \sinh c_2 \cosh(a_2-b_2) \tag{Ad} \\[4pt] &+ \left(\cosh c-\cosh(a-b)\right) \cdot \cosh(a_2+b_2)\;\sinh c_2 \\[6pt] = &\phantom{+\;}\sinh c_2 \left(\; \begin{array}{c} (\cosh a\cosh b - \cosh c)\;( \cosh(a_2-b_2) - \cosh(a_2+b_2)) \tag{Ae} \\[2pt] + \sinh a \sinh b\;( \cosh( a_2 - b_2 ) + \cosh(a_2 + b_2 ) ) \end{array} \;\right) \\[4pt] = &\phantom{+\;}\sinh c_2 \left(\; \begin{array}{c} -2 \sinh a_2 \sinh b_2 (\cosh a\cosh b - \cosh c) \\[2pt] + 2 \sinh a \sinh b \cosh a_2 \cosh b_2 \end{array} \;\right) \tag{Af} \\[4pt] = &\phantom{+\;}2 \sinh c_2 \left(\; \begin{array}{c} \sinh a_2 \sinh b_2 (\cosh c - \cosh a\cosh b ) \\[2pt] + 4 \sinh a_2 \sinh b_2 \cosh^2 a_2 \cosh^2 b_2 \end{array} \;\right) \tag{Ag} \\[4pt] = &\phantom{+\;}2 \sinh a_2 \sinh b_2 \sinh c_2 \left(\; \cosh c - \cosh a\cosh b + ( \cosh a + 1 )( \cosh b + 1 ) \;\right) \tag{Ah} \\[4pt] = &\phantom{+\;}2 \sinh a_2 \sinh b_2 \sinh c_2 \left(\; 1 + \cosh a + \cosh b + \cosh c \;\right) \tag{Ai} \end{align}$$

  • $\begingroup$ can you work out the bit on how you get to (cosh(a+b)-cosh(c))/(2 sinh a sinhb) and the next steps? $\endgroup$
    – Willemien
    Commented Oct 9, 2015 at 9:38
  • $\begingroup$ Use the identity $$\cosh x-\cosh y \equiv 2\sinh(x_2+y_2)\sinh(x_2-y_2)$$ in the expressions for $\cos^2\gamma_2$ and $\sin^2\gamma_2$. Reducing the "$s$" expression is somewhat too involved for a comment; I'll put an addendum to the answer. $\endgroup$
    – Blue
    Commented Oct 9, 2015 at 10:04
  • $\begingroup$ Thanks, can you also number the equations (makes commenting easier) $\endgroup$
    – Willemien
    Commented Oct 9, 2015 at 12:26

The above formula is proved in Wilson Stothers' Cabri Pages

Alternative formula for the area of a right-angled hyperbolic triangle
If the hyperbolic triangle ABC has a right angle at C, then its area D is given by $$\tan\left(\frac{D}{2}\right) = \tanh\left(\frac{a}{2}\right)\tanh\left(\frac{b}{2}\right)$$ Although it is quite easy to prove this directly, we may as well use Heron's formula.
Since the triangle is right-angled at C, $\cosh(c) = \cosh(a)\cosh(b)$, i.e. $\gamma = \alpha\beta$.
Then $\Delta^2 = 1-\alpha^2-\beta^2-\alpha^2\beta^2+2\alpha^2\beta^2 = (\alpha^2-1)(\beta^2-1) = (\sinh(a)\sinh(b))^2$.
Also, $1+\alpha+\beta+\gamma = 1+\alpha+\beta+\alpha\beta = (\alpha+1)(\beta+1) = (\cosh(a)+1)(\cosh(b)+1)$.
Now, Heron's formula gives $\tan\left(\frac{D}{2}\right) = \left(\frac{\sinh(a)}{\cosh(a)+1}\right)\left(\frac{\sinh(b)}{\cosh(b)+1}\right)$.
The result follows as $\frac{\sinh(x)}{\cosh(x)+1} = \tanh\left(\frac{x}{2}\right)$.

also found the formula in Hyperbolic plane geometry revisited. It divides the area of a general hyperbolic triangles into two right-angled hyperbolic triangles.

The area of the triangle: \begin{gathered} T:=2 \delta=\pi-(\alpha+\beta+\gamma) . \\ \tan \frac{T}{2}=\left(\tanh \frac{a_1}{2}+\tanh \frac{a_1}{2}\right) \tanh \frac{m_a}{2}, \end{gathered} where $m_a$ is the height of the triangle corresponding to $A$ and $a_1, a_2$ are the signed lengths of the segments into which the foot point of the height divides the side $B C$.


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