# 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?

• FYI: This answer of mine lists the sines and cosines of multiples of $3^\circ$.
– Blue
Mar 17, 2018 at 11:29
• I just wrote an answer but deleted it. You'd better specify that you don't want to get roots of complex number involved otherwise the answer will be trivialized as $\cos \pi/n, n\in\mathbb Z$ can be always be written as complex roots of $\pm 1$. Mar 19, 2018 at 7:28

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).

• This happens to be one of the ways the golden ratio shows up when you're dealing with regular pentagons. Mar 17, 2018 at 12:12
• "if and only if $p$ is a Fermat prime" .......holy cow!! this just blew my mind. Mar 18, 2018 at 21:01
• The "only if" direction I think was proved by Wantzel, not Gauss. And a detail: The special $p=2$ is of the form $2^{2^k}+1$ only if we allow $k=-\infty$, and usually it is not included in the Fermat primes (whether to do that is certainly just a matter of definition). Mar 18, 2018 at 23:30

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):

$$y \;=\; \frac{1}{4}\left( \,\frac{(1 + i \sqrt{3}) \sqrt{4 + 4 i \sqrt{3}}}{2} + \frac{2(1 - i \sqrt{3})}{ \sqrt{4 + 4 i \sqrt{3}}} \right)$$

WA verifies that $y - \sin \pi/18 = 0$ indeed.

• "(albeit complex radicals since it's casus irreducibilis)." - in fact, one turns this case on its head by expressing the roots of the cubic in terms of trigonometric functions, which avoid the need to deal with complex numbers. Mar 17, 2018 at 12:14
• Can you spell out the complex number? It'd be nice to see it explicitly. Mar 17, 2018 at 15:29
• @Mitch Done and edited into the answer.
– dxiv
Mar 17, 2018 at 19:32
• @J.M.isnotamathematician Very true, though in a way not all that different from solving the quadratic using trig substitutions. In the end, it boils down to polynomial identities for multiples of the angle such as $\,\cos nx = T_n(\cos x)\,$, which "happen" to be solvable by radicals for $\,n=2,3\,$.
– dxiv
Mar 18, 2018 at 2:39
• Right, and one could see this as an elementary version of what happens with quintics and higher degree polynomials: in those cases, one replaces "trigonometric function" with "(hyper)elliptic function", which are functions that in a sense generalize the trigonometric functions. Mar 18, 2018 at 6:13

I find it curious that the Wikipedia table has one entry the doesn't have the minimum depth of surds. Here is yet another table of cosines in increments of $3°$. $$\begin{array}{r|c}\theta&\cos\theta\\\hline 0°&1\\ 3°&\frac1{16}\left(\sqrt6-\sqrt2\right)\left(\sqrt5-1\right)+\frac18\left(\sqrt3+1\right)\sqrt{5+\sqrt5}\\ 6°&\frac18\sqrt3\left(\sqrt5+1\right)+\frac18\sqrt{10-2\sqrt5}\\ 9°&\frac18\sqrt2\left(\sqrt5+1\right)+\frac14\sqrt{5-\sqrt5}\\ 12°&\frac18\left(\sqrt5-1\right)+\frac18\sqrt3\sqrt{10+2\sqrt5}\\ 15°&\frac14\left(\sqrt6+\sqrt2\right)\\ 18°&\frac14\sqrt{10+2\sqrt5}\\ 21°&\frac1{16}\left(\sqrt6+\sqrt2\right)\left(\sqrt5+1\right)+\frac18\left(\sqrt3-1\right)\sqrt{5-\sqrt5}\\ 24°&\frac18\left(\sqrt5+1\right)+\frac18\sqrt3\sqrt{10-2\sqrt5}\\ 27°&\frac18\sqrt2\left(\sqrt5-1\right)+\frac14\sqrt{5+\sqrt5}\\ 30°&\frac12\sqrt3\\ 33°&-\frac1{16}\left(\sqrt6-\sqrt2\right)\left(\sqrt5-1\right)+\frac18\left(\sqrt3+1\right)\sqrt{5+\sqrt5}\\ 36°&\frac14\left(\sqrt5+1\right)\\ 39°&\frac1{16}\left(\sqrt6-\sqrt2\right)\left(\sqrt5+1\right)+\frac18\left(\sqrt3+1\right)\sqrt{5-\sqrt5}\\ 42°&\frac18\sqrt3\left(\sqrt5-1\right)+\frac18\sqrt{10+2\sqrt5}\\ 45°&\frac12\sqrt2\\ 48°&-\frac18\left(\sqrt5-1\right)+\frac18\sqrt3\sqrt{10+2\sqrt5}\\ 51°&\frac1{16}\left(\sqrt6+\sqrt2\right)\left(\sqrt5+1\right)-\frac18\left(\sqrt3-1\right)\sqrt{5-\sqrt5}\\ 54°&\frac14\sqrt{10-2\sqrt5}\\ 57°&\frac1{16}\left(\sqrt6+\sqrt2\right)\left(\sqrt5-1\right)+\frac18\left(\sqrt3-1\right)\sqrt{5+\sqrt5}\\ 60°&\frac12\\ 63°&-\frac18\sqrt2\left(\sqrt5-1\right)+\frac14\sqrt{5+\sqrt5}\\ 66°&\frac18\sqrt3\left(\sqrt5+1\right)-\frac18\sqrt{10-2\sqrt5}\\ 69°&-\frac1{16}\left(\sqrt6-\sqrt2\right)\left(\sqrt5+1\right)+\frac18\left(\sqrt3+1\right)\sqrt{5-\sqrt5}\\ 72°&\frac14\left(\sqrt5-1\right)\\ 75°&\frac14\left(\sqrt6-\sqrt2\right)\\ 78°&-\frac18\sqrt3\left(\sqrt5-1\right)+\frac18\sqrt{10+2\sqrt5}\\ 81°&\frac18\sqrt2\left(\sqrt5+1\right)-\frac14\sqrt{5-\sqrt5}\\ 84°&-\frac18\left(\sqrt5+1\right)+\frac18\sqrt3\sqrt{10-2\sqrt5}\\ 87°&\frac1{16}\left(\sqrt6+\sqrt2\right)\left(\sqrt5-1\right)-\frac18\left(\sqrt3-1\right)\sqrt{5+\sqrt5}\\ 90°&0 \end{array}$$ I used Wolfram alpha to check the Mathjax expressions.

EDIT: A brief explanation about the table: Once one has solved $$\frac{\cos2\theta+\cos\theta}{\cos\theta+1}=0$$ And $$\frac{\cos3\theta+\cos2\theta}{\cos\theta+1}=0$$ For $\cos\frac{\pi}3$ and $\cos\frac{\pi}5$ respectively and obtained $\cos\frac{\pi}4$ by bisection one can get the other trig functions one needs by angle sum formulas and the Pythagorean theorem. Then the Diophantine systems $$\frac x{60}=\frac a3+\frac b4+\frac c5$$ were solved with $|a|\le1$ and $|c|\le2$: $$\begin{array}{r|rrr}x&a&b&c\\\hline 0&0&0&0\\ 1&-1&3&-2\\ 2&1&-2&1\\ 3&0&1&-1\\ 4&-1&0&2\\ 5&1&-1&0\\ 6&0&2&-2\\ 7&-1&1&1\\ 8&1&0&-1\\ 9&0&-1&2\\ 10&-1&2&0\\ 11&1&1&-2\\ 12&0&0&1\\ 13&-1&3&-1\\ 14&1&-2&2\\ 15&0&1&0\\ \end{array}$$ At this point $\cos\left(\frac{\pi a}3+\frac{\pi b}4\right)$ and $\sin\left(\frac{\pi a}3+\frac{\pi b}4\right)$ were determined and finally $\cos\frac{\pi x}{60}$ and $\sin\frac{\pi x}{60}$.

• +1. As I commented to the question, a previous answer of mine also has a list of sines and cosines for multiples of $3^\circ$. There, I'd worked from this list by Scott Surgent to seek a unified form. Interestingly, Surgent's expressions are mostly three-deep radicals, while yours are at most two, which confirms a suspicion I'd had that it should always be possible to escape Surgent's outer-most roots. Now, I wonder: What's the unified form of your versions of the values?
– Blue
Mar 18, 2018 at 22:49
• @Blue I'm not sure what you mean about unified form, but I edited in an explanation of how the table was generated. The link from the deleted post has a similar form to mine besides including angles at $1°$ increments. I have been debating whether to post construction of $\cos\frac{\pi}{11}$ either as an edit or a separate answer. What do you think? Mar 18, 2018 at 23:25
• "unified form" as in my table, where all the value expressions themselves (not the angles to which they correspond) are given by a single formula with four integer parameters and a pair of sign choices. Presumably, your list should have the same kind of underlying structure. (That $1^\circ$ list by Parent is laudable, but pretty impractical as a reference ... and the typesetting gives me a headache. :) There, too, one wonders if there's a way to bring order to the chaos.) As for $\cos(\pi/11)$: I think that would warrant a separate answer.
– Blue
Mar 18, 2018 at 23:48
• I've added a "unified form" for these values to my previous answer.
– Blue
Mar 19, 2018 at 18:41

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 $sin(15)=sin\widehat{BAC}=\frac{BC}{CA}=\frac{-3+2\sqrt{3}}{\sqrt{{24-12\sqrt{3}}}}=\frac{-3+2\sqrt{3}}{3\sqrt{2}-\sqrt{6}}=\frac{\sqrt{6}-\sqrt{2}}{4}$.

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.

With absolutely no proof at all, the next least complicated answer is for $\cos( \frac{\pi}{17})$:

$$\cos \frac{\pi}{17} = \frac{ \sqrt{15 + \sqrt{17} + \sqrt{34 - 2\sqrt{17}} + \sqrt{68 + 12\sqrt{17} - 4\sqrt{170 + 38\sqrt{17}}}} }{4\sqrt{2}}$$

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.

• I think you mean "because algebraic numbers form a field". Mar 17, 2018 at 19:09
• When you say "if and only if $x$ is rational", you seem to mean (as your subsequent discussion shows) "if and only if $x/\pi$ is rational". Mar 17, 2018 at 19:13
• @EricTowers Fixed, thanks! Mar 18, 2018 at 13:27
• Hmm... When you switched to "then $\sin x^\circ$ can be expressed using a radical expression if and only if $x/\pi$ is rational." you either needed the degrees symbol or the division by $\pi$ -- both gets you to another error. If $\sin x$, then $x/\pi$ must be rational. If $\sin x^\circ$, then $x$ must be rational (or $x^\circ \cdot \frac{\pi}{180^\circ} \cdot \frac{1}{\pi}$ must be rational). Mar 18, 2018 at 16:29

For angles like $\frac{\pi}{11}$ the construction is kinda hard, so we'll take a couple of shortcuts to provide hopefully the right answer, but lacking rigorous proof. If you consider the set of values $$\left\{2\cos\frac{2\pi}{11},2\cos\frac{6\pi}{11},2\cos\frac{4\pi}{11},2\cos\frac{10\pi}{11},2\cos\frac{8\pi}{11}\right\}$$ We can cycle through them in order by tripling the angle at each step. Thus the operation $R$ that triples the angle is a realization of the group $\mathbb{Z}_5$. Now it strains my sense of mathematical abstraction to think of an operation that multiplies by functional composition as an addition, so instead I will think in terms of the isomorphic point group of rotations of multiples of $72°$ about the $z$-axis, $C_5$. Here is its character table: $$\begin{array}{c|ccccc}C_5&E&C_5&C_5^2&C_5^3&C_5^4\\\hline A&1&1&1&1&1\\ E_{1+}&1&\omega&\omega^2&\omega^3&\omega^4\\ E_{2+}&1&\omega^2&\omega^4&\omega&\omega^3\\ E_{2-}&1&\omega^3&\omega&\omega^4&\omega^2\\ E_{1-}&1&\omega^4&\omega^3&\omega^2&\omega\end{array}$$ Where $\omega=\exp\left(\frac{2\pi i}5\right)=\frac{\sqrt5-1}4+i\frac{\sqrt{10+2\sqrt5}}4$. Using the operator $$\sum_{R\in C_5}D_{jk}^{(\mu)}\left(R^{-1}\right)R$$ That projects into the $k^{\text{th}}$ partner of the $\mu^{\text{th}}$ irreducible representation of $C_5$ we can generate $5$ functions that transform as the irreducible representations in order: \begin{align}\theta_0&=2\cos\frac{2\pi}{11}+2\cos\frac{6\pi}{11}+2\cos\frac{4\pi}{11}+2\cos\frac{10\pi}{11}+2\cos\frac{8\pi}{11}\\ \theta_1&=2\cos\frac{2\pi}{11}+2\omega^4\cos\frac{6\pi}{11}+2\omega^3\cos\frac{4\pi}{11}+2\omega^2\cos\frac{10\pi}{11}+2\omega\cos\frac{8\pi}{11}\\ \theta_2&=2\cos\frac{2\pi}{11}+2\omega^3\cos\frac{6\pi}{11}+2\omega\cos\frac{4\pi}{11}+2\omega^4\cos\frac{10\pi}{11}+2\omega^2\cos\frac{8\pi}{11}\\ \theta_3&=2\cos\frac{2\pi}{11}+2\omega^2\cos\frac{6\pi}{11}+2\omega^4\cos\frac{4\pi}{11}+2\omega\cos\frac{10\pi}{11}+2\omega^3\cos\frac{8\pi}{11}\\ \theta_4&=2\cos\frac{2\pi}{11}+2\omega\cos\frac{6\pi}{11}+2\omega^2\cos\frac{4\pi}{11}+2\omega^3\cos\frac{10\pi}{11}+2\omega^4\cos\frac{8\pi}{11}\end{align} This transform is invertible: \begin{align}2\cos\frac{2\pi}{11}&=\frac15\left(\theta_0+\theta_1+\theta_2+\theta_3+\theta_4\right)\\ 2\cos\frac{6\pi}{11}&=\frac15\left(\theta_0+\omega\theta_1+\omega^2\theta_2+\omega^3\theta_3+\omega^4\theta_4\right)\\ 2\cos\frac{4\pi}{11}&=\frac15\left(\theta_0+\omega^2\theta_1+\omega^4\theta_2+\omega\theta_3+\omega^3\theta_4\right)\\ 2\cos\frac{10\pi}{11}&=\frac15\left(\theta_0+\omega^3\theta_1+\omega\theta_2+\omega^4\theta_3+\omega^2\theta_4\right)\\ 2\cos\frac{8\pi}{11}&=\frac15\left(\theta_0+\omega^4\theta_1+\omega^3\theta_2+\omega^2\theta_3+\omega\theta_4\right)\end{align} Now comes the nonrigorous part: we are going to compute values numerically and assume that our results are correct and exact: \begin{align}\theta_0&=-1\\ \theta_1\theta_4&=11\\ \theta_2\theta_3&=11\\ \theta_1^5+\theta_4^5+\theta_2^5+\theta_3^5&=-979\\ (\theta_1^5+\theta_4^5-\theta_2^5-\theta_3^5)^2&=378125\end{align} Given that we know which signs to take from our previous numerical results, we have $$\theta_1^5+\theta_4^5=\theta_1^5+\frac{11^5}{\theta_1^5}=\frac{11}2\left(-89+25\sqrt5\right)$$ $$\theta_1^{10}-\frac{11}2\left(-89+25\sqrt5\right)\theta_1^5+11^5=0$$ $$\theta_1^5=\frac{11}4\left(-89+25\sqrt5+5i\sqrt{410+178\sqrt5}\right)$$ Now, we have to be careful because when we take $5^{\text{th}}$ roots the phase won't normally be correct, so $$\theta_1=\omega^4\sqrt{\frac{11}4\left(-89+25\sqrt5+5i\sqrt{410+178\sqrt5}\right)}$$ Similarly we can work out $$\theta_2=\omega\sqrt{\frac{11}4\left(-89-25\sqrt5-5i\sqrt{410-178\sqrt5}\right)}$$ $$\theta_3=\omega^4\sqrt{\frac{11}4\left(-89-25\sqrt5+5i\sqrt{410-178\sqrt5}\right)}$$ $$\theta_4=\omega\sqrt{\frac{11}4\left(-89+25\sqrt5-5i\sqrt{410+178\sqrt5}\right)}$$ Since $\cos\frac{\pi}{11}=-\cos\frac{10\pi}{11}$ we have \begin{align}\cos\frac{\pi}{11}&=-\frac1{10}\left\{-1+\frac{-\sqrt5-1+i\sqrt{10-2\sqrt5}}4\sqrt{\frac{11}4\left(-89+25\sqrt5+5i\sqrt{410+178\sqrt5}\right)}\right.\\ &\quad+\left.\frac{-\sqrt5-1+i\sqrt{10-2\sqrt5}}4\sqrt{\frac{11}4\left(-89-25\sqrt5-5i\sqrt{410-178\sqrt5}\right)}\right.\\ &\quad+\left.\frac{-\sqrt5-1-i\sqrt{10-2\sqrt5}}4\sqrt{\frac{11}4\left(-89-25\sqrt5+5i\sqrt{410-178\sqrt5}\right)}\right.\\ &\quad+\left.\frac{-\sqrt5-1-i\sqrt{10-2\sqrt5}}4\sqrt{\frac{11}4\left(-89+25\sqrt5-5i\sqrt{410+178\sqrt5}\right)}\right\}\end{align} I was hoping to show also my construction of $\sin\frac{\pi}{11}$ but it's too late now. Maybe tomorrow.

EDIT: Time for $\sin\frac{\pi}{11}$. This time we have the set of values $$\left\{2\cos\frac{\pi}{22},2\cos\frac{5\pi}{22},2\cos\frac{19\pi}{22},2\cos\frac{7\pi}{22},2\cos\frac{9\pi}{22}\right\}$$ Which we can cycle through via the operation $S$ that quintuples angles this time. As before we construct \begin{align}\phi_0&=2\cos\frac{\pi}{22}+2\cos\frac{5\pi}{22}+2\cos\frac{19\pi}{22}+2\cos\frac{7\pi}{22}+2\cos\frac{9\pi}{22}\\ \phi_1&=2\cos\frac{\pi}{22}+2\omega^4\cos\frac{5\pi}{22}+2\omega^3\cos\frac{19\pi}{22}+2\omega^2\cos\frac{7\pi}{22}+2\omega\cos\frac{9\pi}{22}\\ \phi_2&=2\cos\frac{\pi}{22}+2\omega^3\cos\frac{5\pi}{22}+2\omega\cos\frac{19\pi}{22}+2\omega^4\cos\frac{7\pi}{22}+2\omega^2\cos\frac{9\pi}{22}\\ \phi_3&=2\cos\frac{\pi}{22}+2\omega^2\cos\frac{5\pi}{22}+2\omega^4\cos\frac{19\pi}{22}+2\omega\cos\frac{7\pi}{22}+2\omega^3\cos\frac{9\pi}{22}\\ \phi_4&=2\cos\frac{\pi}{22}+2\omega\cos\frac{5\pi}{22}+2\omega^2\cos\frac{19\pi}{22}+2\omega^3\cos\frac{7\pi}{22}+2\omega^4\cos\frac{9\pi}{22}\end{align} By now the reader knows the drill about inverting this DFT to recover cosines. Again we compute, albeit approximately and with a leap of faith: \begin{align}\phi_0^2&=11\\ \phi_1\phi_4&=11\\ \phi_2\phi_3&=11\\ \left(\phi_1^5+\phi_4^5+\phi_2^5+\phi_3^5\right)^2&=130691=11\cdot109^2\\ \left(\phi_1^5+\phi_4^5-\phi_2^5-\phi_3^5\right)^2&=34375=55\cdot25^2\end{align} I suppose we could have established the above results by observing that the primaries were sums of $220^{\text{th}}$ roots of unity and collected results in $220$ buckets of integers, but we didn't. Again we solve as far as $$\phi_1^5+\phi_4^5=\phi_1^5+\frac{11^5}{\phi_1^5}=\frac{109\sqrt{11}+25\sqrt{55}}2$$ $$\phi_1^{10}-\frac{109\sqrt{11}+25\sqrt{55}}2\phi_1^5+11^5=0$$ $$\phi_1^5=\frac{\sqrt{11}}4\left(109+25\sqrt5+5i\sqrt{8770-218\sqrt5}\right)$$ Solving for the other variables and being careful about phase when we take fifth roots we find \begin{align}\phi_0&=\sqrt{11}\\ \phi_1&=\sqrt{\frac{\sqrt{11}}4\left(109+25\sqrt5+5i\sqrt{8770-218\sqrt5}\right)}\\ \phi_2&=\omega^4\sqrt{\frac{\sqrt{11}}4\left(109-25\sqrt5-5i\sqrt{8770+218\sqrt5}\right)}\\ \phi_3&=\omega\sqrt{\frac{\sqrt{11}}4\left(109-25\sqrt5+5i\sqrt{8770+218\sqrt5}\right)}\\ \phi_4&=\sqrt{\frac{\sqrt{11}}4\left(109+25\sqrt5-5i\sqrt{8770-218\sqrt5}\right)}\end{align} Applying the inverse DFT, \begin{align}\sin\frac{\pi}{11}&=\cos\frac{9\pi}{22}=\frac1{10}\left(\phi_0+\omega^4\phi_1+\omega^3\phi_2+\omega^2\phi_3+\omega\phi_4\right)\\ &=\frac1{10}\left\{\sqrt{11}+\frac{\sqrt5-1-i\sqrt{10+2\sqrt5}}4\sqrt{\frac{\sqrt{11}}4\left(109+25\sqrt5+5i\sqrt{8770-218\sqrt5}\right)}\right.\\ &\quad+\left.\frac{-\sqrt5-1+i\sqrt{10-2\sqrt5}}4\sqrt{\frac{\sqrt{11}}4\left(109-25\sqrt5-5i\sqrt{8770+218\sqrt5}\right)}\right.\\ &\quad+\left.\frac{-\sqrt5-1-i\sqrt{10-2\sqrt5}}4\sqrt{\frac{\sqrt{11}}4\left(109-25\sqrt5+5i\sqrt{8770+218\sqrt5}\right)}\right.\\ &\quad+\left.\frac{\sqrt5-1+i\sqrt{10+2\sqrt5}}4\sqrt{\frac{\sqrt{11}}4\left(109+25\sqrt5-5i\sqrt{8770-218\sqrt5}\right)}\right\}\end{align}

• this is horrifying
– Rchn
Mar 19, 2018 at 15:17
• If we are allowed to take roots of complex numbers the question would be trivialized. If we are not allowed then this is not a valid answer. See my comment on the question. Mar 19, 2018 at 22:50
• @WeijunZhou Does taking roots mean the roots you want or the principal roots? I think this is the kind of stuff the OP wanted to see, and it's kind of hard to find it lying around on the web. Even though I use the trigonometric and hyperbolic forms for solution to the cubic equation because you're going to have to use trig and inverse trig functions to take the cube root of a complex number, the algebraic form, as above, shows we can construct a regular $11$-gon given an angle pentasector. Also rather like Rader FFT method. I thought you should construct $\cos\frac{\pi}7$ rather than delete. Mar 19, 2018 at 23:42
• My deleted answer is exactly about $\cos \pi/7$, presumably using the Cardano formula. Even if you claim that you only take the principal root $\exp(2i\pi/n)$, You can construct the other root by $1/\exp(2i\pi/n)$ and then construct $\cos(2\pi/n)$ from it. Mar 19, 2018 at 23:50
• @Rchn this is beautiful. Oct 14, 2018 at 9:05

A while ago, I accidentally discovered this one (angles in degree): $$\sin 37=\sin 67 \sqrt{\frac{1 - \sin 16}{3/2-\sin 16+\sqrt{2}\sin 23\sqrt{1-\sin 16}}}$$

• which apparently boils down to $\sin^2 67-\sin^2 37=\frac 14+\sin 37\cos 67$. Nothing special... Mar 17, 2018 at 11:40

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 $72^{\circ}-36^{\circ}-72^{\circ}$ 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.