Are there any "nonstandard" special angles for which trig functions yield radical expressions? Everyone learns about the two "special" right triangles at some point in their math education—the $45-45-90$ and $30-60-90$ triangles—for which we can calculate exact trig function outputs. But are there others?
To be specific, are there any values of $y$ and $x$ such that: 


*

*$y=\sin(x)$; 

*$x$ (in degrees) is not an integer multiple of $30$ or $45$; 

*$x$ and $y$ can both be written as radical expressions? By radical expression, I mean any finite formula involving only integers, addition/subtraction, multiplication/division, and $n$th roots. [Note that I require $x$ also be a radical expression so that we can't simply say "$\arcsin(1/3)$" or something like that as a possible value of $x$, which would make the question trivial.]
If yes, are they all known and is there a straightforward way to generate them?
If no, what's the proof?
 A: A while ago, I accidentally discovered this one (angles in degree):
$$\sin 37=\sin 67 \sqrt{\frac{1 - \sin 16}{3/2-\sin 16+\sqrt{2}\sin 23\sqrt{1-\sin 16}}}$$
A: Actually, you can construct many of them using the Half-Angle Formula
$$
\sin(\frac{x}{2})= \pm \sqrt{\frac{1-\cos(x)}{2}}
$$
, the sign depending on the quadrant $x$ is located. Note that the Pythagorean Identity $\cos^2(x)+\sin^2(x)=1$ implies that $\cos(x)$ is a radical expression (as defined in the OP) if and only $\sin(x)$ is.
Therefore, starting with one example $y=\sin(x)$ that satisfies the problem (for example, $x=45$ degrees and $y=\frac{\sqrt{2}}{2}$), your can construct a sequence of new examples. 
In addition, because of the Addition Formulas like
$$\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$$
, you can obtain a new pair of examples $(x,y)$ form examples $(x_1,y_1)$ and $(x_2,y_2)$. In particular, if $x$ yields a radical expression for $y=\sin(x)$, then so does $kx$ for any positive integer $k$.
Remark: This generalizes TheSimpliFire's answer.
A: A $72^{\circ}-36^{\circ}-72^{\circ}$ isosceles triangle gives $$\sin 18^{\circ}=\frac{\sqrt 5-1}{4}$$

$\Delta ABC$ is an isosceles triangle. 
$AB=AC=a, \angle A=32^{\circ}, AD\bot BC, CE$ is bisector of $\angle C$ and $EF\bot AC$. We observe that $BC=EC=EA=x$.
Now, $\Delta ABC\sim CEB.$ Hence,
$$\frac{AB}{CE}=\frac{BC}{EB}\\
\frac ax=\frac{x}{a-x}\implies x^2+ax-a^2=0\\
\frac xa=\frac{\sqrt 5-1}{2}\\
\sin \angle BAD=\frac{BD}{AB}\\
\sin 18^{\circ}=\frac{\frac x2}{a}=\frac{\sqrt 5-1}{4}$$
Similarly $\cos 36^{\circ}$ can be found from this triangle.
A: There is
$$\cos\frac{\pi}5=\frac{\sqrt5+1}4$$
and similar for cosines and sines of multiples of this. Gauss proved that
one can find expressions for $\cos \pi/p$ involving iterated square roots
where $p$ is prime if and only if $p$ is a Fermat prime (of form $2^{2^k}+1$), so for $p=2$, $3$, $5$, $17$, $257$ and $65537$ (but to date no others are known).
A: Note that $\sin(3x) = 3 \sin(x) - 4 \sin^3(x)$ so you can always "trisect an angle in radicals" since the cubic equation is solvable by radicals.
For example, taking $\,3x = 30^\circ\,$ gives the cubic in $\,y = \sin 10^\circ\,$ as $\,8y^3 - 6y + 1 = 0\,$ where the root $\,y\,$ can be expressed by radicals (albeit complex radicals since it's a casus irreducibilis).

[ EDIT ]   As requested in a comment, the following is the explicit form of the solution for the sample case above (where the radicals denote the principal value of the fractional powers):

$$
y \;=\; \frac{1}{4}\left( \,\frac{(1 + i \sqrt{3}) \sqrt[3]{4 + 4 i \sqrt{3}}}{2} + \frac{2(1 - i \sqrt{3})}{ \sqrt[3]{4 + 4 i \sqrt{3}}} \right)
$$
WA verifies that $y - \sin \pi/18 = 0$ indeed.
A: Yes, a $15-75-90$ triangle may be the one you want.

Assume we have a right $\Delta ABC$ with $\widehat{BAC}=15^0;\widehat{ABC}=90^0;\widehat{ACB}=75^0$.
Put an extra point $D$ like above so that $B,C,D$ are collinear and $AC$ is the angle bisector of $\widehat{DAB}$, this means $\widehat{DAB}=30^0;\widehat{BDA}=60^0$.
Let $DB=a$. Then the special right triangle $\Delta ABD$ will have $AD=2a$ and $AB=\sqrt{3}a$.
Because $AC$ is the angle bisector of $\widehat{DAB}$, we have $\frac{CB}{CD}=\frac{AB}{AD}=\frac{\sqrt{3}}{2}$.
We have this set of equations: ${\begin{cases}DB=CB+CD=a\\\frac{CB}{CD}=\frac{\sqrt{3}}{2}\end{cases}} \Rightarrow {\begin{cases}CD=\left(4-2\sqrt{3}\right)a\\CB=\left(-3+2\sqrt{3}\right)a\end{cases}}$
Apply the Pythagorean theorem: $CA=\sqrt{AB^2+BC^2}=\sqrt{(\sqrt{3a})^2+((-3+2\sqrt{3})a)^2}=\sqrt{(24-12\sqrt{3})a^2}=\sqrt{24-12\sqrt{3}}a$
We conclude that $sin(15)=sin\widehat{BAC}=\frac{BC}{CA}=\frac{-3+2\sqrt{3}}{\sqrt{{24-12\sqrt{3}}}}=\frac{-3+2\sqrt{3}}{3\sqrt{2}-\sqrt{6}}=\frac{\sqrt{6}-\sqrt{2}}{4}$.
A: There is also $$\begin{align}\tan\frac\pi8&=-1+\sqrt2\\\tan\frac{3\pi}8&=1+\sqrt2\\\tan\frac{5\pi}8&=-1-\sqrt2\\\tan\frac{7\pi}8&=1-\sqrt2\end{align}$$ For proofs of the first two see here.
A: With absolutely no proof at all, the next least complicated answer is for $\cos( \frac{\pi}{17})$:
$$\cos \frac{\pi}{17} = 
\frac{
  \sqrt{15 + 
  \sqrt{17} + 
  \sqrt{34 - 2\sqrt{17}} + 
  \sqrt{68 + 12\sqrt{17} - 4\sqrt{170 + 38\sqrt{17}}}}
}{4\sqrt{2}} $$
(from Ron Knott, Exact Trig Function Values)
A: For the record: I do not know if "radical expression" is a commonly used phrase; someone people comment telling me if it is.
If $x$ is algebraic (which every radical expression is, because algebraic numbers form a field), then $\sin{x^\circ}$ can be expressed using a radical expression if and only if $x$ is rational.
To prove this, for any $z\in\mathbb{C}$, we have
$$\sin{z}=\frac{e^{iz}-e^{-iz}}{2i}=\frac{(-1)^{\frac{z}{\pi}}-\frac{1}{(-1)^{\frac{z}{\pi}}}}{2i},$$
which is a radical expression if and only if $(-1)^{\frac{z}{\pi}}$ is a radical expression (proof left as an exercise; let me know if you want a hint).
This uses $z$ in radians, so if $x$ is an angle measure in degrees, then $\sin{x^{\circ}}=\sin{\pi \frac{x}{180}}$ is a radical expression if and only if $(-1)^{\frac{x}{180}}$ is a radical expression.
We will now use the Gelfond–Schneider theorem, which states that if $a,b\in\mathbb{R}$ are algebraic with $a\notin\{0,1\}$ and $b\notin\mathbb{Q}$, then $a^b$ is transcendental.  In this case, because $(-1)^2=1\ne-1$, we have that if $x$ is irrational, then $(-1)^{\frac{x}{180}}$ is transcendental and therefore not algebraic (and therefore not a radical expression).
On the other hand, if $x$ is rational, then 
$$\sin{x^\circ}=\sin{\pi\frac{x}{180}}=\frac{(-1)^{\frac{x}{180}}-\frac{1}{(-1)^{\frac{x}{180}}}}{2i}$$
is an expression of $\sin{x}$ in radicals.  This might not be quite what you are looking for; for the answer to when this can be expressed in a form I think you are looking for see this.
