Why is a pure tone sinusoidal? Why not any other shape to represent a single frequency (like triangular waves)?
I haven't been able to find a glorious "easy to understand answer", but I did find that sinusoidal displacement of a particle produces a sinusoidal velocity and acceleration. Could this explain why this is the ideal or natural shape of a motion and waves in nature, as well as it's use in analysis techniques like Fourier transforms?
 A: Sinusoids (sines and cosines) are the eigenfunctions of the wave equation. That is if you look for a set of functions $f_{i,\omega}$ for which it is true that
$$ \frac{\partial^2 f_{i,\omega}}{\partial x} \pm \frac{1}{c^2} \frac{\partial^2 f_{i,\omega}}{\partial t} = \lambda f_{i,\omega} \;, $$
for some real number $\lambda$, then all the answers would be some combinations of $\sin(kx \mp \omega t)$ and $\cos (kx \mp \omega t)$.1
This is a pretty special property. For any other periodic functions (such as triangle waves or square waves) there are no qualifying solutions at all. However, you can write all those non-qualifying solutions as a sum of the ones that do qualify. Letting $F$ be a stand in for any non-qualifying periodic solution, then 
$$
F = \sum_i \int \mathrm{d}\omega\, c_{i,\omega} \, f_{i,\omega} \;, 
$$
for some selected set of coefficient $c_{i,\omega}$s.2
So, long and mathematical story short: sinusoidal waves have the special properties of having unique frequency and a being usable to compose all periodic functions.

1 I included the $i$ subscript on $f_{i,\omega}$ so that we could distinguish $\sin$ and $\cos$. It turns out that $\lambda = k^2 - \frac{\omega^2}{c^2}$, and that by choosing $k$ (for a given $\omega$) so that $\lambda = 0$ you force the wave speed to be $c$ and have a description of a physical wave that matches the conditions of your apparatus.
2 There is even a way to find the correct $c_{i,\omega}$s from $F$, but writing it down here wouldn't actually explain any more about why the sinusoids are special.
