Why does the Pythagorean Theorem have its simple form only in Euclidean geometry? Below are the right-angled forms of the Pythagorean Theorem in elliptic, Euclidean, and hyperbolic geometry, respectively.
$$\cos\left(\frac{c}{R}\right) = \cos\left(\frac{a}{R}\right)\cos\left(\frac{b}{R}\right)$$
$$c^2 = a^2 + b^2$$
$$\cosh c = \cosh a \; \cosh b$$
To me, it looks like the Euclidean version is the simplest. No trig functions (and only one trig function in the general case). Also, it is the only version to have anything squared; the other two have $f(c)=f(a)f(b)$ whereas the Euclidean version is $f(c)=f(a)+f(b)$. Furthermore, there are plenty of fairly simple proofs of the Pythagorean Theorem in Euclidean geometry, but I would guess that this is not the case for elliptic and hyperbolic geometries. Why is this so?
My hunch is that the Parallel Postulate is the key to this answer. However, I don't see the connection between having exactly one parallel line and whatever it is that makes the formula turn out so nicely.
 A: You can even generalize the law of cosines.

$$
\cos\left(\frac{z}{r}\right) =
    \cos\left(\frac{x}{r}\right) \cos\left(\frac{y}{r}\right)
    + \cos(\phi) \sin\left(\frac{x}{r}\right) \sin\left(\frac{y}{r}\right).
\tag 1
$$

What we have is
$$
\begin{array}{rcl}
r^2 > 0 &\rightarrow& \textrm{spherical}\\
&& \cos\left(\frac{z}{r}\right) =
    \cos\left(\frac{x}{r}\right) \cos\left(\frac{y}{r}\right)
    + \cos(\phi) \sin\left(\frac{x}{r}\right) \sin\left(\frac{y}{r}\right)\\\\
r^2 < 0 &\rightarrow& \textrm{hyperbolic}\\
&&
\cosh\left(\frac{z}{r}\right) =
    \cosh\left(\frac{x}{r}\right) \cosh\left(\frac{y}{r}\right)
    - \cos(\phi) \sinh\left(\frac{x}{r}\right) \sinh\left(\frac{y}{r}\right)\\\\
\lim_{\displaystyle r^2 \rightarrow \infty} &\rightarrow& \textrm{flat or Euclidean}\\
&& z^2 = x^2 + y^2 - 2 \cos(\phi) x y
\end{array}
$$

Write it out...
$$
\begin{array}{rcl}
\displaystyle
    \left( 1 - \frac{1}{2} \left( \frac{z}{r} \right)^2 + \cdots \right)
&=&
\displaystyle
    \left( 1 - \frac{1}{2} \left( \frac{x}{r} \right)^2 + \cdots \right)
    \left( 1 - \frac{1}{2} \left( \frac{y}{r} \right)^2 + \cdots \right)\\
&& \displaystyle \hspace{2em} + \cos(\phi)
    \left( \frac{x}{r} + \cdots \right)
    \left( \frac{y}{r} + \cdots \right)
\end{array}
$$
So you end up with
$$
z^2 = x^2 + y^2 - 2 \cos(\phi) x y.
$$
