Is there any deep reason that special trig angles are a sequence of roots of integers? We all are familiar with certain special values of sin and cos, e.g. $\sin(30^\circ)=0.5$, $\sin(45^\circ)=\frac{1}{\sqrt{2}}$ etc. In high school I found these values difficult to remember until I noticed they could be reformulated as this pattern:
$$\sin(0^\circ)=\frac{\sqrt{0}}{2}$$
$$\sin(30^\circ)=\frac{\sqrt{1}}{2}$$
$$\sin(45^\circ)=\frac{\sqrt{2}}{2}$$
$$\sin(60^\circ)=\frac{\sqrt{3}}{2}$$
$$\sin(90^\circ)=\frac{\sqrt{4}}{2}$$
This has been bugging me for years ever since. I know to expect there to be some deep cause of patterns when I see them, but I have no idea what causes these common special angle values to be half-roots-of-integers, or whether this pattern is just a special case of a more general notion of special trig angles.
What's the explanation? Why should this be so?
 A: The intervals between the angles are not uniform.  Somehow it's mathematical coincidence:
Since $\sin (90^{\circ}-\theta)=\sqrt{1-\sin^{2} \theta}$,
$$\sin 0^{\circ}=\frac{\sqrt{0}}{2} \iff \sin 90^{\circ}=\frac{\sqrt{4}}{2}$$
$$\sin 30^{\circ}=\frac{\sqrt{1}}{2} \iff \sin 60^{\circ}=\frac{\sqrt{3}}{2}$$
We might invent a mnemonic in this way:
$$\sin 35.26^{\circ}=\dfrac{\sqrt{1}}{\sqrt{3}}$$
$$\sin 54.74^{\circ}=\dfrac{\sqrt{2}}{\sqrt{3}}$$
These angles relate to tetrahedron and are special angles in another way.
Comparing the $\color{green}{\textbf{fitting curve}}$ with $\color{red}{\boldsymbol{\sin \dfrac{\pi x}{12}}}$ below:

A: Unsatisfying explanation: $\sin: [0,\pi/2]\to [0,1]$ is a monotonically increasing homeomorphism, so you'll always be able to find values which solve $\sin(a)=b$ for $b\in[0,1]$ and if you pick $b$ to be a sequence like $\frac{\sqrt{k}}{n}$ where $k$ runs from $0$ to $n^2$, you'll see the same sort of mirrored progression in the angles, as $\sin^2(x)+\sin^2(\pi/2-x)=1$.
To me, the real gem here is that rational multiples of $\pi$ give algebraic numbers under $\sin$ and $\cos$ (and therefore under $\tan,\sec,\csc,\cot$ where appropriately defined). Here's why: 
Suppose $\theta$ is a rational multiple of $\pi$. We can write it as a rational multiple of $2\pi$: $\theta=\frac{2p\pi}{q}$. By De Moivre's formula, we have that $(\cos\theta+i\sin\theta)^q=\cos(q\theta)+i\sin(q\theta)=1$, so $\cos\theta+i\sin\theta$ is algebraic, and so each of $\cos\theta$ and $\sin\theta$ are as well (if $a+bi$ is algebraic, then $a-bi$ is also algebraic, as algebraicness is preserved under field automorphisms, and so $(a+bi)+(a-bi)=2a$ is algebraic as well, and therefore since algebraic numbers form a field, we have that $a=\frac{2a}{2}$ and $b=\frac{a+bi-a}{i}$ are algebraic).
A: Let's make a table.
$$\begin{array}{ccccc|l}\frac{\sqrt 0}{2}&\frac{\sqrt 1}{2}&\frac{\sqrt 2}{2}&\frac{\sqrt 3}{2}&\frac{\sqrt 4}{2}&\text{sin}\\\hline0&30&45&60&90&\text{angle (deg)}
\\0&2&3&4&6&\text{multiple of 15}\\
1&0&1&0&1&\text{divisible by 45?}\\
1&1&0&1&1&\text{divisible by 30?}\\\end{array}$$
By periodicity this is a combination of the cyclic groups $\frac{360}{45} = 8$ and $\frac{360}{30} = 12$ considering rotation as the generator. This means we will stay in the set $\pm\left\{\frac{\sqrt 0}{2},\frac{\sqrt 1}{2},\frac{\sqrt 2}{2},\frac{\sqrt 3}{2},\frac{\sqrt 4}{2}\right\}$ if starting from 0, subset $\pm\left\{\frac{\sqrt 0}{2},\frac{\sqrt 1}{2},\frac{\sqrt 3}{2},\frac{\sqrt 4}{2}\right\}$ if 30 degrees and $\pm\left\{\frac{\sqrt 0}{2},\frac{\sqrt 2}{2},\frac{\sqrt 4}{2}\right\}$ if 45 degrees. Number theoretic every integer times 15 deg which is a multiple of 2 or 3. And with a rotation of 15 degrees we can get a rather peculiar overlap.
