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.

• I like this...! – JohnWO Jul 26 '15 at 1:12
• In fairness, other things are simpler in non-euclidean geometries. The area of a triangle, for instance is determined by the angles (if total curvature is known). And the formula for area is linear in the angles! Can't get any simpler... – lulu Jul 26 '15 at 1:39
• The disparity grows in higher dimensions. If $W$ is the "hypotenuse-face" of a right-corner tetrahedron, and $X$, $Y$, $Z$ the "leg-faces", then Euclidean space has $$|W|^2 = |X|^2+|Y|^2+|Z|^2$$ (where "$|\cdot|$" is face area), but non-Euclidean space has $$\cos\frac{|W|}{2}=\cos\frac{|X|}{2}\cos\frac{|Y|}{2}\cos\frac{|Z|}{2}\pm \sin\frac{|X|}{2}\sin\frac{|Y|}{2}\sin\frac{|Z|}{2}$$("$-$" for hyperbolic, "$+$" for spherical). Indeed, in dimensions $4+$, Euclid always has $$|\text{hyp}|^2=|\text{leg}|^2+|\text{leg}|^2+\cdots+|\text{leg}|^2$$ The non-Euclidean counterparts are ... unknown! – Blue Jul 26 '15 at 1:45

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 $$z^2 = x^2 + y^2 - 2 \cos(\phi) x y.$$

• I quite like this explanation. I wish I could give another +1 for teaching me that $\cos(ix) = \cosh(x)$; that's a very intriguing and fascinating link between the elliptical and hyperbolic geometries. – El'endia Starman Jul 26 '15 at 2:18
• The parallel postulate you mentioned in the other post has indeed a lot to do with this ;) – johannesvalks Jul 26 '15 at 2:22
• I feel compelled to mention the Laws of Cosines for tetrahedra. Writing "$X_2$" for $X/2$, and "$\angle XY$" for the dihedral angle between faces $X$ and $Y$, Euclid has $$W^2=X^2+Y^2+Z^2-2XY\cos\angle XY-2YZ\cos\angle YZ-2ZX\cos\angle ZX$$ Hyperbolic/spherical space has \begin{align}\cos W_2&=\cos X_2\cos Y_2\cos Z_2\pm\sin X_2\sin Y_2\sin Z_2\sqrt{S}\\&+\cos X_2\sin Y_2\sin Z_2\cos\angle YZ \\&+\sin X_2\cos Y_2\sin Z_2\cos\angle ZX\\&+\sin X_2\sin Y_2\cos Z_2\cos\angle XY\end{align} where $$S:=1-2\cos\angle XY\cos\angle YZ\cos\angle ZX-\cos^2\angle XY-\cos^2\angle YZ-\cos^2 \angle ZX$$ – Blue Jul 26 '15 at 3:46
• @Blue, thanks for this comment. – johannesvalks Jul 26 '15 at 14:34