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I'm trying to prove the following for the above Lambert quadrilateral: $\cos \beta = \sinh b \sinh \alpha$. I tried tracing a diagonal from vertex $v(a,b)$ to $v(\alpha,c)$and using hyperbolic sine and cosine laws, but calculations got huge and probably that's the wrong way.

I've already proved the relations in right triangle $(a,b,c,\alpha,\beta,\pi/2)$:

$\cosh c = \cot \alpha \cot \beta$

$\sin \alpha = \frac{\sinh a}{\sinh c} $

$\cos \alpha = \sin \beta \cosh a $

$\cos \alpha = \frac{ \tanh b}{ \tanh c}$

$\cosh c = \cosh a \cosh b$

$\tan \alpha = \frac{ \tanh a}{ \sinh b}$


  • $\begingroup$ Any help with how to reduce the Lambert quad to triangles will be appreciated. $\endgroup$ – user286485 Oct 7 '16 at 20:38

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Sometimes, you just have to push through these kinds of manipulations. This one isn't really so bad.

By the Law of Cosines for Angles:

$$\cos\psi = -\cos\theta^\prime \cos\phi^\prime + \sin\theta^\prime \sin\phi^\prime \cosh r \tag{1}$$

Since $\theta + \theta^\prime = \phi + \phi^\prime = \pi/2$, and since $r$ is the hypotenuse of a triangle with legs $p$ and $q$ ...

$$\cos\psi = -\sin\theta \sin\phi + \cos\theta \cos\phi \cdot \cosh p \cosh q \tag{2}$$

Expressing the trig functions of $\theta$ and $\phi$ in terms of $p$ and $q$, and simplifying with some hyperbolic trig identities ...

$$\begin{align} \cos\psi &= - \frac{\sinh p}{\sinh r} \frac{\sinh q}{\sinh r} + \frac{\tanh p}{\tanh r} \frac{\tanh q}{\tanh r} \cdot \cosh p \cosh q \tag{3} \\[6pt] &= - \frac{\sinh p \sinh q}{\sinh^2 r} + \frac{\sinh p \sinh q}{\tanh^2 r} \tag{4} \\[6pt] &= \left( -\operatorname{csch}^2 r + \coth^2 r \right) \sinh p \sinh q \tag{5} \\[6pt] &= 1 \cdot \sinh p \sinh q \quad\square \tag{6} \end{align}$$


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