Are there any "nonstandard" special angles for which trig functions yield radical expressions?

Question

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?

Answer 1

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).

Answer 2

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).

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[ 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.

Answer 3

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 $\sin\frac{\pi x}{60}$.

Answer 4

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

Answer 5

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.

Answer 6

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)

Answer 7

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.

Answer 8

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 $$\theta_1=\omega^4\sqrt[5]{\frac{11}4\left(-89+25\sqrt5+5i\sqrt{410+178\sqrt5}\right)}$$ Similarly we can work out $$\theta_2=\omega\sqrt[5]{\frac{11}4\left(-89-25\sqrt5-5i\sqrt{410-178\sqrt5}\right)}$$ $$\theta_3=\omega^4\sqrt[5]{\frac{11}4\left(-89-25\sqrt5+5i\sqrt{410-178\sqrt5}\right)}$$ $$\theta_4=\omega\sqrt[5]{\frac{11}4\left(-89+25\sqrt5-5i\sqrt{410+178\sqrt5}\right)}$$ Since $\cos\frac{\pi}{11}=-\cos\frac{10\pi}{11}$ we have $$\begin{align}\cos\frac{\pi}{11}&=-\frac1{10}\left\{-1+\frac{-\sqrt5-1+i\sqrt{10-2\sqrt5}}4\sqrt[5]{\frac{11}4\left(-89+25\sqrt5+5i\sqrt{410+178\sqrt5}\right)}\right.\\ &\quad+\left.\frac{-\sqrt5-1+i\sqrt{10-2\sqrt5}}4\sqrt[5]{\frac{11}4\left(-89-25\sqrt5-5i\sqrt{410-178\sqrt5}\right)}\right.\\ &\quad+\left.\frac{-\sqrt5-1-i\sqrt{10-2\sqrt5}}4\sqrt[5]{\frac{11}4\left(-89-25\sqrt5+5i\sqrt{410-178\sqrt5}\right)}\right.\\ &\quad+\left.\frac{-\sqrt5-1-i\sqrt{10-2\sqrt5}}4\sqrt[5]{\frac{11}4\left(-89+25\sqrt5-5i\sqrt{410+178\sqrt5}\right)}\right\}\end{align}$$ 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 $$\begin{align}\phi_0&=\sqrt{11}\\ \phi_1&=\sqrt[5]{\frac{\sqrt{11}}4\left(109+25\sqrt5+5i\sqrt{8770-218\sqrt5}\right)}\\ \phi_2&=\omega^4\sqrt[5]{\frac{\sqrt{11}}4\left(109-25\sqrt5-5i\sqrt{8770+218\sqrt5}\right)}\\ \phi_3&=\omega\sqrt[5]{\frac{\sqrt{11}}4\left(109-25\sqrt5+5i\sqrt{8770+218\sqrt5}\right)}\\ \phi_4&=\sqrt[5]{\frac{\sqrt{11}}4\left(109+25\sqrt5-5i\sqrt{8770-218\sqrt5}\right)}\end{align}$$ Applying the inverse DFT, $$\begin{align}\sin\frac{\pi}{11}&=\cos\frac{9\pi}{22}=\frac1{10}\left(\phi_0+\omega^4\phi_1+\omega^3\phi_2+\omega^2\phi_3+\omega\phi_4\right)\\ &=\frac1{10}\left\{\sqrt{11}+\frac{\sqrt5-1-i\sqrt{10+2\sqrt5}}4\sqrt[5]{\frac{\sqrt{11}}4\left(109+25\sqrt5+5i\sqrt{8770-218\sqrt5}\right)}\right.\\ &\quad+\left.\frac{-\sqrt5-1+i\sqrt{10-2\sqrt5}}4\sqrt[5]{\frac{\sqrt{11}}4\left(109-25\sqrt5-5i\sqrt{8770+218\sqrt5}\right)}\right.\\ &\quad+\left.\frac{-\sqrt5-1-i\sqrt{10-2\sqrt5}}4\sqrt[5]{\frac{\sqrt{11}}4\left(109-25\sqrt5+5i\sqrt{8770+218\sqrt5}\right)}\right.\\ &\quad+\left.\frac{\sqrt5-1+i\sqrt{10+2\sqrt5}}4\sqrt[5]{\frac{\sqrt{11}}4\left(109+25\sqrt5-5i\sqrt{8770-218\sqrt5}\right)}\right\}\end{align}$$

Answer 9

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

Answer 10

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.

Answer 11

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.