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Looking through the Wolfram MathWorld sections on trigonometric angles gets us to certain values of $\cos{(\theta)}$ and $\sin{(\theta)}$. Most of these angles can be expressed as radicals of complex numbers. They also can be done with polynomial roots of degree $3$ or less or through easier ways, such as through the half angle formulas, for example to get $\sin{\left(\frac{\pi}{12}\right)}$, the half angle formula could be used on $\sin{\left(\frac{\pi}{3}\right)}$ twice or once on $\sin{\left(\frac{\pi}{6}\right)}$.

Some of these, however cannot be expressed very neatly with the website, and most certainly on others, not giving their exact values in terms if . Some of these are the cosine and sine of $\theta$ where $\theta = \frac{a \pi}{n}$, where $a,n \in\Bbb Z$. To start off with, let us use $\theta' = \frac{\pi}{n}$where ∈{11,13,19,21,23,27,29,31} You can also do any in the list above or some other prime number denominators.

I understand that this may be a little bit difficult to evaluate, but if these values could be written in radical form of complex numbers such as the example on the website of $\frac{\pi}{9}$, please tell me.

You can also tell me the values of cosine and sine of $\frac {aθ'}{n}$ if you chοοse. It would be preffered to be typed out in the fully simplified form in its entirety.

This means if there is an $(-1)^{\frac pq}$ in the equivalent expression where $p,q \in \Bbb Z$, then it should be expanded into the “irreducible” expression of radicals and Complex numbers. If this is too complex, the RootReduce[TrigToRadicals[Sin[Pi/n]]] and Developer`TrigToRadicals[Sin[Pi/n]] commands can be typed into it.

I suspect the Wolfram software is required as my attempts to use the first one have not worked in WolframAlpha in the App. You can also use substitutions like in the $\frac {\pi}{17}$ example in the webpage below. It may also be a good idea to post a link with the answer on maple, using the aforementioned commands in WolframAlpha, or your own idea. Can $\sin(\pi/25)$ be expressed in radicals https://mathworld.wolfram.com/topics/TrigonometricIdentities.html

Thanks! You are contributing to the world of Trigonometry!

Тyma Gaidash
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  • You refer to "Wolfram Alpha sections". Please give the URL for this so we can know what you found there. – Somos Mar 30 '21 at 18:39
  • You can ask WolframAlpha to find the roots the minimal polynomial of, say, $\sin(\pi/11)$. However, since your denominators of interest ($11$, $13$, $19$, $23$) are not Fermat primes, the roots (and thus the sines) won't be expressible using radicals. – Blue Mar 30 '21 at 18:53
  • @Somos. Sorry about that. I have just linked the sections above Enjoy – Тyma Gaidash Mar 30 '21 at 18:53
  • These are, of course, algebraic, and can be expressed using radicals. Why (or why not) is this not acceptable:$$\cos\frac{\pi}{11} = \frac{(-1)^{1/11}+(-1)^{-1/11}}{2}$$I used the "principal value" for the $11$th root. Of course we do not expect these can be written in radicals without the use of complex numbers. – GEdgar Mar 30 '21 at 18:53
  • @Blue. For one example, on the Pi/13, website that says the value “has a cyclic Galois group, and so x, and hence sin(pi/13), can be expressed in terms of radicals (of complex numbers)”. this means that the value may not be expressible using only real numbers, square roots, and cube roots, but still may be with complex numbers and higher surd roots. Correct me if I am wrong. – Тyma Gaidash Mar 30 '21 at 19:00
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    @Blue I apologize for overriding your edit, however I tried to incorporate more “MathJax” into the question as well as a bit of grammar correction. I will try to apply some of your edits. – Тyma Gaidash Mar 30 '21 at 19:03
  • @TymaGaidash: "this means that the value [...] still may be with complex numbers and higher surd root" ... Well, there is that. My geometric inclinations often have me lose interest if values aren't constructible (ie, expressible with real radicals). :) – Blue Mar 30 '21 at 19:05
  • @GEdgar as said in the question, it would be appreciated that the answers would be written in the exact expression of the value with radicals incorporated with complex numbers or in a way that could be expressed using them. For example the infinitely nested square root of the π/18 example instead of using the half angle formula on the π/9 example with complex numbers and radicals. n=9,n=18 – Тyma Gaidash Mar 30 '21 at 19:39
  • @Babado Thank you for the edits. Yours, Blue’s and a couple of mine really made the question a more professionally stated one and made it easier to comprehend. – Тyma Gaidash Mar 30 '21 at 19:49
  • You're welcome. Try to get used to using mathjax – Babado Mar 30 '21 at 20:41
  • Before i may ponder starting to type an answer, the question wants explicit radical expressions for the sine and cosine values of all rational multiples of $\pi$, where the denominator of that rational factor is a prime $p$ covering all values in the list ${11,13,19,21,23,27,29,31}$?! (I am missing an explicit question, the OP comes also with no effort to compute anything...) Is it enough for instance to show the value for $$\cos\frac\pi{13}\ ?$$ Or the complete computational path has to be given? Possibly also all subfields in the tower of the field $\Bbb Q(\zeta_{13})$? – dan_fulea May 06 '21 at 17:31
  • ... Please make clear which is the question and to what extent an answer will be accepted. Covering all primes in the above list is of course a nice wish, a knowledge which is nice to have, but making the computations... and then typing them... Is computer support allowed? May i use for instance sage? Is it enough to provide some code for some typical cases? (Of course, different primes dividing $(p-1)$ make the types differ in the nature, but the investigation follows parallel lines...) When investigating $21$, why not looking for $7$ directly...? – dan_fulea May 06 '21 at 17:38
  • @dan_fulea, in paragraph 2, the question states “ To start off with, let us use” referring to the value of a=1 in the numerator to start off with. This means that the other values are not be be computed, they would probably looks similar if they were irreducible anyways, until the numerator of 1 angles are found out. Also, “ If this is too complex, the RootReduce[TrigToRadicals[Sin[Pi/n]]] and Developer`TrigToRadicals[Sin[Pi/n]] commands can be typed into it” means you can totally use computer support. – Тyma Gaidash May 06 '21 at 17:49
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    There is not too much difference between computations for different values of $a$, when the denominator is kept constant, but there is a huge difference when passing from $13$ (which is fairly simple to decide) to $23$. For $p=23$, the number field involved has dimension $22$ over $\Bbb Q$, so understanding $$\zeta_{23}=\exp\left(\frac 1{23}\cdot 2\pi i\right)$$and all its cousins using radical expressions depends on the fact that some subfield of dimension $11$ is a radical extensions of $\Bbb Q$. The decision is not simple in this case... Computers can use Galois data - but this is data... – dan_fulea May 06 '21 at 17:56
  • The problem is not really a trigonometry problem, it is rather a problem related to the explicit computation (using only radicals) of cyclotomic units. Doing this for $13$ is one job, doing it for some other value like $31$ (involving the complicated prime $5$ dividing $(31-1)$) or $23$ (involving eleven, really complicated case), a different job. The cited sentence "To start..." is unfortunately not a question. Stating a clear question makes it easier to start an answer, and focus on a clear matter. – dan_fulea May 06 '21 at 18:02

2 Answers2

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This is an answer showing the limitation of the "computational wish", taken instead for the lack of a question.

The question is rather a question related to computations in number fields. (Which are field extensions of finite degree of the base field $\Bbb Q$.) We are especially dealing with cyclotomic fields. There is already a notion of a "radical extension" related to extensions of fields, but this notion also covers $\sqrt[n]{\pm 1}$, and using this notion, the problem of expressing the cyclotomic units by (such) radicals is already solved.

So we are searching for more explicit formulas. A natural wish is to use - when possible - only the radicals of some order computed for a very specific real number. This is too specific, below there will be some more or less explicit formulas. For instance, we obtain a formula for $$ \cos\frac \pi{13} + i\sin\frac \pi{13} $$ which is not the cheap "$\displaystyle\sqrt[13]{-1}$", but rather a formula involving some more explicit radicals of smaller degree, of the shape $C+\sqrt[3]{A+B}+\sqrt[3]{A-B}$, where $A,B,C$ are complex numbers written explicitly in terms of radicals and algebraic (field) operations. It is still a task to isolate the real and the imaginary part of $\sqrt[3]{A+B}$, but well, i will not do it here, hoping that the progress made is enough.

Same qualitative situation applies to the other cases, however these cases come with "higher" complications.


The case $p=13$:

I will start with the simplest case, and "compute" the sine and cosine for $\pi/13$. For this, let $u$ be the complex number $$ u = \exp \frac {\pi i}{13} = \cos \frac {\pi}{13} +i \sin \frac {\pi}{13} = \cos \frac {2\pi}{26} +i \sin \frac {2\pi}{26} \ . $$ Here are some facts about this number. It is a cyclotomic unit of degree $23$ It satisfies the equation $u^{13}=\exp(\pi i)=-1$, so it is a root of the polynomial of degree $13-1=12=2^2\cdot 3$: $$ q = X^{12} - X^{11} + X^{10} - X^9 + X^8 - X^7 + X^6 - X^5 + X^4 - X^3 + X^2 - X + 1\ . $$ The Galois group of the related cyclotomic field is cyclic and abelian, so we expect a tower of subfields corresponding to the subgroups of $\Bbb Z/12$. They are joined via edges in a parallel way as the edges of the graph of divisors of $12$, an edge being drawn if there is a divisibility relation. (The following investigation can be done using either the divisibility path $1|2|4|12$ or the divisibility path $1|3|6|12$ or... My choice will be the one involving third roots as late as possible.)

Let $K=K_{12}=\Bbb Q(u)$ be the field generated by $u$. Then we have a tower of subfields... $$ \require{AMScd} \color{darkgreen}{ \begin{CD} K=K_{12}=\Bbb Q(u)\\ @V V :3 V\\ K_4=\Bbb Q(a) @. \qquad\text{ where }a=\frac14\Big(1\color{brown}{\pm}\sqrt{ 13}\color{blue}{\pm}\sqrt{2(13\color{brown}{\pm} 3\sqrt {13})}\ \Big) \\ @VV :2 V\\ K_2=\Bbb Q(\sqrt{13})\\ @VV :2 V\\ \Bbb Q \end{CD}} $$ Being an extension of degree three, $K:K_4$, it is radical, and we can "compute" $u$ in terms of $a$. The minimal polynomial of $a$ over $\Bbb Q$ is $x^4 - x^3 + 2x^2 + 4x + 3$. We are not fixing the two signs in the definition of $a$, and compute algebraically $u$ as the root of the polynomial with coefficients in $K_4$: $$ f = X^3 - aX^2 + \frac 13(a^3 + 2a + 1)X + 1\ . $$ (Already the idea of applying the formulas of Del Ferro, Tartaglia, Cardano explicitly makes me feel some pain in the finger.) Then $u-\frac 13a$ satisfies $$ g = x^3 + \frac 13(a^3 - a^2 + 2a + 3)x + \frac1{27}(a^3 - 3a + 18)\ . $$ We are searching for two numbers, $s,t$ with $-3st=\frac 13(a^3 - a^2 + 2a + 3)$ and $s^3+t^3=-\frac1{27}(a^3 - 3a + 18)$. We know the sum and product of the numbers $s^3$, and $t^3$, so these two numbers are $$ \begin{aligned} s^3, t^3 &= -\frac1{54}(a^3 - 3a + 18) \pm \frac 12\sqrt{\frac 1{27}(a^3 - a + 11)} \\ &= \frac 1{108}\Big(-26 + 4\sqrt{13} + i\sqrt{26 + 6\sqrt{13}}\Big) \ \pm\ \frac 1{36}\sqrt{6(13-3\sqrt{13})} \\ &=A\pm B \ . \end{aligned} $$ The above results are easy, when displayed, well, getting them... Now the three possible values for $\displaystyle v = u-\frac 13a$ are finally the following ones: $$ \varepsilon^k s + \bar\varepsilon^k t = \varepsilon^k\sqrt[3]{A + B} + \bar\varepsilon^k\sqrt[3]{A - B}\ ,\qquad k=0,1,2\ . $$ Here $\varepsilon$ is a primitive third root of unity. And please do not ask now for the explicit form of the real part of these numbers... (The third roots should be chosen in each line so that the corresponding product $-3st$ has the needed value.)

It is clear, that we obtain formulas using only radicals and algebraic operations for the roots of the polynomials $g$ and $f$ above.


Let us see if the numerical values of the above expressions have any chance...

a = ( 1 + sqrt(13) + i*sqrt( 2*(13 + 3*sqrt(13)) ) ) / 4
print( f"a = {a} \n  ~ {a.n()}" ) 
print( f"The minimal polynomial of a is {a.minpoly()}\n" )

We introduce an algebraic version of the above a, and related fields.

K4.<a> = NumberField(x^4 - x^3 + 2x^2 + 4x + 3) R.<X> = PolynomialRing(K4) L.<u> = K4.extension(X^3 - aX^2 + (1/3a^3 + 2/3a + 1)X + 1) print( f"u is a root of the polynomial:\n{u.minpoly()}\n" ) print( f"Then we have u^13 = {u**13}" )

var('x'); for ac in a.complex_embeddings(): print(f"a is numerically ~ {ac}")

P = (x^3 -ac*x^2 + (1/3*ac^3 + 2/3*ac + 1)*x + 1)

uc_values = P.roots(ring=CC, multiplicities=False)
for uc in uc_values:
    k = round( real( (log(uc)/i/pi*13).n() ), 6 ) 
    print(f&quot;    u is numerically {uc} ... angle -&gt; {k:g} pi / 13&quot;)

And we obtain first:

a = 1/4*sqrt(13) + 1/4*I*sqrt(6*sqrt(13) + 26) + 1/4 
  ~ 1.15138781886600 + 1.72542218842201*I
The minimal polynomial of a is x^4 - x^3 + 2*x^2 + 4*x + 3

u is a root of the polynomial: x^3 - ax^2 + (1/3a^3 + 2/3a + 1)x + 1

Then we have u^13 = -1

confirming the fact that all values of $u$ algebraically found by the above receipt are satisfying $u^{13}=-1$. Then we have using numerical approximations for $a$ corresponding roots $u$, written also numerically, and guess the corresponding angle...

a is numerically ~ -0.651387818865997 - 0.522415803456408*I
    u is numerically -0.885456025653210 - 0.464723172043769*I ... angle -> -11 pi / 13
    u is numerically -0.120536680255323 - 0.992708874098054*I ... angle -> -7 pi / 13
    u is numerically 0.354604887042536 + 0.935016242685415*I ... angle -> 5 pi / 13
a is numerically ~ -0.651387818865997 + 0.522415803456408*I
    u is numerically -0.885456025653210 + 0.464723172043769*I ... angle -> 11 pi / 13
    u is numerically -0.120536680255323 + 0.992708874098054*I ... angle -> 7 pi / 13
    u is numerically 0.354604887042536 - 0.935016242685415*I ... angle -> -5 pi / 13
a is numerically ~ 1.15138781886600 - 1.72542218842201*I
    u is numerically -0.568064746731156 - 0.822983865893656*I ... angle -> -9 pi / 13
    u is numerically 0.748510748171100 - 0.663122658240795*I ... angle -> -3 pi / 13
    u is numerically 0.970941817426053 - 0.239315664287557*I ... angle -> -1 pi / 13
a is numerically ~ 1.15138781886600 + 1.72542218842201*I
    u is numerically -0.568064746731156 + 0.822983865893656*I ... angle -> 9 pi / 13
    u is numerically 0.748510748171100 + 0.663122658240795*I ... angle -> 3 pi / 13
    u is numerically 0.970941817426053 + 0.239315664287557*I ... angle -> 1 pi / 13

Now that we also have some formulas involving only radicals (over $\Bbb C$) and algebraic operations, let us implement some of the required angles... The following simple code involves only the mentioned operations and can be easily translated in some other CAS...

w = sqrt(13)

a = ( 1 + w + isqrt( 26 + 6w ) ) / 4 s3 = ( -26 + 4w + isqrt( 26 + 6w ) ) / 108 + 1/36 sqrt( 6(13 - 3w) ) t3 = ( -26 + 4w + isqrt( 26 + 6w ) ) / 108 - 1/36 sqrt( 6(13 - 3w) )

st = -1/9 * (a^3 - a^2 + 2a + 3) s = s3 ^ (1/3) # extract cubic root #t = t3 ^ (1/3) # extract cubic root ... but better insure st = st t = st / s

eps = (-1 + isqrt(3))/2 u1 = a/3 + s + t u2 = a/3 + eps s + eps^2 * t u3 = a/3 + eps^2 * s + eps * t

for u in (u1, u2, u3): k = ZZ(round( real( log(u.n()) / (ipi) 13 ), 6 )) print(f"u is numerically {u.n()}") print(f" cos({k} pi / 13) ~ {cos(kpi / 13).n()}") print(f" sin({k} pi / 13) ~ {sin(kpi / 13).n()}")

And we obtain:

u is numerically 0.970941817426052 + 0.239315664287557*I
    cos(1 pi / 13) ~ 0.970941817426052
    sin(1 pi / 13) ~ 0.239315664287558
u is numerically -0.568064746731156 + 0.822983865893657*I
    cos(9 pi / 13) ~ -0.568064746731156
    sin(9 pi / 13) ~ 0.822983865893656
u is numerically 0.748510748171102 + 0.663122658240795*I
    cos(3 pi / 13) ~ 0.748510748171101
    sin(3 pi / 13) ~ 0.663122658240795

So if we must write a formula, it is: $$ \color{blue}{ \begin{aligned} &w := \sqrt{13}\ ,\\ & \cos\frac\pi{13} + i\sin\frac\pi{13} \\ &\qquad= \frac 1{12}\Big( 1 + w + i\sqrt{26 + 6w} \Big) \\ &\qquad\qquad +\left(\ \frac 1{108} ( -26 + 4w + i\sqrt{26 + 6w} )\ +\ \frac 1{36}\sqrt{6(13-3w)}\ \right)^{1/3} \\ &\qquad\qquad +\left(\ \frac 1{108} ( -26 + 4w + i\sqrt{26 + 6w} )\ -\ \frac 1{36}\sqrt{6(13-3w)}\ \right)^{1/3} \ . \end{aligned}} $$ Here, the two cubic radicals are taken from complex numbers, and they have to be inter-coordinated, they have to match in the sense, that their product

  • defined up to some third root of unity - should be a known quantity, $-\frac 19(a^3 - a^2 + 2a + 3)$. A choice of the first cubic root determines thus the second cubic root, and the value of the above expression shows the sine and the cosine of $\pi/13$, or of $3\pi/3$, or of $9\pi/13$.

Here is a further possible path to obtain the "simpler" cosine values, and then by trigonometry the "more complicated" sine values for the angles with denominator $13$. We blindly use sage to get the minimal polynomial of the algebraic number $c=2\cos(\pi/13)$. Then $c$ determines a field of degree six over $\Bbb Q$. We split this extension again in two parts. The splitting can be either the one corresponding to $6=2\cdot 3$, or the one corresponding to $3\cdot 2$. (At any rate there are two strategies.) Depending on strategy (and its meaning, which is still to be defined / anticipated), we will get expressions involving

  • either radicals of expressions involving cubic radicals of "base expressions",
  • or cubic radicals of expressions involving radicals of "base expressions", where the "base expressions" differ from case to case, and may also show radicals.

The picture would be $$ \require{AMScd} \color{darkgreen}{ \begin{CD} K=\Bbb Q(c) @. \qquad\text{ where $c$ is a root of }x^6 - x^5 - 5x^4 + 4x^3 + 6x^2 - 3x - 1\in\Bbb Q[x]\\ @V V :2 V\\ K_3=\Bbb Q(g) @. \qquad\text{ where $g$ is a root of }x^3 - x^2 - 4x - 1\in\Bbb Q[x]\\ @VV :3 V\\ \Bbb Q \end{CD}} $$ and the polynomials are extracted from...

p = ( 2*cos(pi/13) ).minpoly()
print(f"c := 2 cos( pi/13 ) is a root of the polynomial:\n    {p} .")

K.<c> = NumberField( p ) L, morphism, _ = K.subfields( degree=3, name='g' )[0] print(L)

We obtain:

c := 2 cos( pi/13 ) is a root of the polynomial:
    x^6 - x^5 - 5*x^4 + 4*x^3 + 6*x^2 - 3*x - 1 .
Number Field in g0 with defining polynomial x^3 - x^2 - 4*x - 1

For the above picture, we have to solve the equation of degree three for $g$, using the formula of Del Ferro, Tartaglia, Cardano, then an equation of degree two with coefficients depending on $g$.

There is also the "other way", the picture would be: $$ \require{AMScd} \color{darkgreen}{ \begin{CD} K=\Bbb Q(c) @. \qquad\text{ where $c$ is a root of } X^3-\frac 12(h+1)X^2 -X +\frac 12(h+3)\in K_2[x]\\ @V V :3 V\\ K_2=\Bbb Q(h) @. \qquad\text{ where $h=\sqrt {13}$}\\ @VV :3 V\\ \Bbb Q \end{CD}} $$ To obtain this picture, we first ask sage for the subfield(s) of degree two, obtain only $K_2=\Bbb Q(h)=\Bbb Q(\sqrt{13})$, then use it, and split the now known minimal polynomial of $c$ over $K_2$:

L.<h> = QuadraticField(13)
RL.<X> = PolynomialRing(L)

factor( X^6 - X^5 - 5X^4 + 4X^3 + 6X^2 - 3X - 1 )

and we obtain the factorization:

(X^3 + (-1/2*h - 1/2)*X^2 - X + 1/2*h + 3/2) * (X^3 + (1/2*h - 1/2)*X^2 - X - 1/2*h + 3/2)

We took one of the factors above. (Note that the free coefficients $\frac 12(h\pm 3)$ are units in the ring of integers of $K_2$.) And the same formula of Del Ferro - Tartaglia - Cardano gives an "explicit formula" involving only (more or less explicit) radicals for $c=2\cos(\pi/13)$.


It is always a pleasure to work in a cyclotomic field.

This concludes the case $p=13$.


The case $p=11$:

We will adapt one of the above strategies only for this case. Let $u$ be the complex number $$ u = \exp \frac {\pi i}{11} = \cos \frac {\pi}{11} +i \sin \frac {\pi}{11} = \cos \frac {2\pi}{22} +i \sin \frac {2\pi}{22} \ . $$ It satisfies the equation $u^{11}=\exp(\pi i)=-1$, so it is a root of the polynomial of degree $10=2\cdot 5$: $$ q = X^{10} - X^9 + X^8 - X^7 + X^6 - X^5 + X^4 - X^3 + X^2 - X + 1\ . $$ It is a cyclotomic unit of degree $22$. The Galois group of the related cyclotomic field is cyclic and abelian, so we expect a tower of subfields corresponding to the subgroups of $\Bbb Z/10$. We have in a picture for $K=K_{10}:=\Bbb Q(u)$, the field generated by $u$, the following tower of subfields... $$ \require{AMScd} \color{darkgreen}{ \begin{CD} K=K_{10}=\Bbb Q(u)\\ @V V :2 V\\ K_5=\Bbb Q(c) @. \qquad\text{ where $c$ is a root of }x^5 - x^4 - 4x^3 + 3x^2 + 3x - 1\in\Bbb Q[x]\\ @VV :5 V\\ \Bbb Q \end{CD}} $$ (Here, i want "first" the equation of degree five.) The variable $c$, one of its complex embeddings, is a posteriori $2\cos(\pi/11)$.

We are again not allowing ourselves to use the expression $\sqrt[11]{-1}$ as a "solution", instead, more explicit radical expressions are searched.

Being a Galois extension of degree five, $K_5:\Bbb Q$, we have informations about the possible structure of its roots.

The following sage code finds the above minimal polynomial for $c=2\cos\frac\pi{11}$:

sage: p = (2*cos(pi/11)).minpoly()
sage: p
x^5 - x^4 - 4*x^3 + 3*x^2 + 3*x - 1
sage: p.discriminant().factor()
11^4

It turns out that the roots of the polynomial $x^5 - x^4 - 4x^3 + 3x^2 + 3x - 1$ have the structural description: $$ \begin{aligned} c_1 &= &&\frac 15(1 + s + t + u + v) &&\color{gray}{=\frac 15(1 + \sqrt[5]S + \sqrt[5]T + \sqrt[5]U + \sqrt[5]V)}\ ,\\ % c_2 &=&&\frac 15(1 + \zeta s + \zeta^2 t + \zeta^3 u + \zeta^4 v) &&\color{gray}{=\frac 15(1 + \zeta \sqrt[5]S + \zeta^2 \sqrt[5]T + \zeta^3 \sqrt[5]U + \zeta^4 \sqrt[5]V)}\ ,\\ % c_3 &=&&\frac 15(1 + \zeta^2 s + \zeta^4 t + \zeta u + \zeta^3 v) &&\color{gray}{=\frac 15(1 + \zeta^2 \sqrt[5]S + \zeta^4 \sqrt[5]T + \zeta \sqrt[5]U + \zeta^3 \sqrt[5]V)}\ ,\\ % c_4 &=&&\frac 15(1 + \zeta^3 s + \zeta t + \zeta^4 u + \zeta^2 v) &&\color{gray}{=\frac 15(1 + \zeta^3 \sqrt[5]S + \zeta \sqrt[5]T + \zeta^4 \sqrt[5]U + \zeta^2 \sqrt[5]V)}\ ,\\ % c_5 &=&&\frac 15(1 + \zeta^4 s + \zeta^3 t + \zeta^2 u + \zeta v) &&\color{gray}{=\frac 15(1 + \zeta^4 \sqrt[5]S + \zeta^3 \sqrt[5]T + \zeta^2 \sqrt[5]U + \zeta \sqrt[5]V)}\ , \end{aligned} $$ or in a compact, structural writing: $$ \begin{bmatrix} c_1\\c_2\\c_3\\c_4\\c_5 \end{bmatrix} = \frac 15 \underbrace{ \begin{bmatrix} 1 & 1 & 1 & 1 & 1\\ 1 & \zeta^{1\cdot 1}& \zeta^{1\cdot 2}& \zeta^{1\cdot 3}& \zeta^{1\cdot 4}\\ 1 & \zeta^{2\cdot 1}& \zeta^{2\cdot 2}& \zeta^{2\cdot 3}& \zeta^{2\cdot 4}\\ 1 & \zeta^{3\cdot 1}& \zeta^{3\cdot 2}& \zeta^{3\cdot 3}& \zeta^{3\cdot 4}\\ 1 & \zeta^{4\cdot 1}& \zeta^{4\cdot 2}& \zeta^{4\cdot 3}& \zeta^{4\cdot 4}\\ \end{bmatrix}}_{=:A(\zeta)} \begin{bmatrix} 1\\s\\t\\u\\v \end{bmatrix} $$ The inverse of the matrix $A(\zeta)$ is up to a factor a matrix of the same shape, in fact, if we formally (Galois) substitute $\zeta^{-1}$ instead of $\zeta$ in $A(\zeta)$, and compute $A(\zeta)A(\zeta^{-1})$, then we obtain five times the unit matrix, so we have direct access to $s,t,u,v$ in the form: $$ \underbrace{ \begin{bmatrix} 1 & 1 & 1 & 1 & 1\\ 1 & \zeta^{-1\cdot 1}& \zeta^{-1\cdot 2}& \zeta^{-1\cdot 3}& \zeta^{-1\cdot 4}\\ 1 & \zeta^{-2\cdot 1}& \zeta^{-2\cdot 2}& \zeta^{-2\cdot 3}& \zeta^{-2\cdot 4}\\ 1 & \zeta^{-3\cdot 1}& \zeta^{-3\cdot 2}& \zeta^{-3\cdot 3}& \zeta^{-3\cdot 4}\\ 1 & \zeta^{-4\cdot 1}& \zeta^{-4\cdot 2}& \zeta^{-4\cdot 3}& \zeta^{-4\cdot 4}\\ \end{bmatrix}}_{=A(\zeta^{-1})} \begin{bmatrix} c_1\\c_2\\c_3\\c_4\\c_5 \end{bmatrix} = \begin{bmatrix} 1\\s\\t\\u\\v \end{bmatrix} $$ So far, we have cheaply complicated the story. Why do we introduce $s,t,u,v$ in a complicated manner, when the conjugated cos-values are that simple?

Here is the reason. Let us consider $S=s^5$, $T=t^5$, $U=u^5$, $V=v^5$. Then $S,T,U,V$ are the roots of the polynomial of degree four: $$ (X^4 - 979X^3 + 467181X^2 - 157668929X + 25937424601)=$$ $$(X^4 - 11\cdot 89 X^3 + 3^3\cdot11^3\cdot13 X^2 - 11^6\cdot 89 X + 11^{10})\ . $$ (All the above depends on a good order of the cosine related values $c_1,c_2,c_3,c_4,c_5$.)

Degree four is good news when formulas involving radicals are needed. And we have an explicit formula for these roots, simpler than the general one - for Galois reasons, since there is no degree $3$ field extension involved. They come in two pairs of conjugated complex numbers: $$ S,T,U,V \qquad=\qquad \frac {11}4(89 \pm 25\sqrt 5)\qquad\boxed{\pm}\qquad i\frac {55}4\sqrt{2(205\mp 89\sqrt 5)}\ , $$ where the signs $\pm$ and $\mp$ correspond in their choice (if the first one is plus, the second one is minus, and conversely), and the sign $\boxed{\pm}$ is chosen independently.

The four numbers $S,T,U,V$ have each the absolute value $11^{5/2}$.

Now we extract the fifth root from each of the above numbers, and obtain the four numbers $s,t,u,v$, each one has the absolute value $11^{1/2}=\sqrt{11}$.

It turns out that $stuv=11^2$. So we have some constraints, after choices and a possible renotation we even have / can take $t=\bar s=11/s$, $v=\bar u=11/u$. Choices of $s,u$ with $su\ne 11$ determine the situation. We will try all possiblities, but only in some cases we will have a match.

Here is the computer check, there are $5^2=25$ cases computed numerically, and $5$ of them lead to wanted cosine values:

a = sqrt(5)

S = 11/4(89 + 25a) + i55/4sqrt(2(205 - 89a)) U = 11/4(89 - 25a) + i55/4sqrt(2(205 + 89a))

s = S.n()^(1/5) t = s.conjugate() u = U.n()^(1/5) v = u.conjugate()

units5 = [ exp(2pii*k/5).n() for k in [0..4] ]

for z5, w5 in cartesian_product([ units5, units5 ]): s1, t1, u1, v1 = sz5, t/z5, uw5, v/w5 c = ( 1 + s1 + t1 + u1 + v1 ).real() / 10. # potential cos value if c < -1 or c > 1: continue angle_denominator = arccos(c) / pi.n() * 11 k = round(angle_denominator, 0)
if abs(angle_denominator - k) < 1e-5: print(f'Radical expression = {c:18.15f} :: cos({k}pi/11) = {cos(k*pi/11).n():18.15f}...')

This delivers:

Radical expression =  0.142314838273285 :: cos(5pi/11) =  0.142314838273285...
Radical expression =  0.654860733945285 :: cos(3pi/11) =  0.654860733945285...
Radical expression = -0.415415013001886 :: cos(7pi/11) = -0.415415013001886...
Radical expression = -0.841253532831181 :: cos(9pi/11) = -0.841253532831181...
Radical expression =  0.959492973614497 :: cos(1pi/11) =  0.959492973614497...

This concludes the case $p=11$.


The case $p=19$:

Let $u$ be the complex number $$ u = \exp \frac {\pi i}{19} = \cos \frac {\pi}{19} +i \sin \frac {\pi}{19} \ . $$ It satisfies the equation $u^{19}=\exp(\pi i)=-1$, so it is a root of the polynomial of degree $18=2\cdot 3 ²$: $$ q = X^{18} - X^{17} + \dots + X^4 - X^3 + X^2 - X + 1\ . $$ The Galois group of the related cyclotomic field is cyclic and abelian, so we expect a tower of subfields corresponding to the subgroups of $\Bbb Z/18$. We expect that $$ c = 2\cos\frac \pi{19} $$ satisfies an equation of degree $9=18/2$. This is obtained via...

sage: c = 2*cos(pi/19)
sage: c.minpoly()
x^9 - x^8 - 8*x^7 + 7*x^6 + 21*x^5 - 15*x^4 - 20*x^3 + 10*x^2 + 5*x - 1

sage: K.<c> = NumberField( (2cos(pi/19)).minpoly() ) sage: L, _, _ = K.subfields(degree=3)[0] sage: L Number Field in c0 with defining polynomial x^3 - x^2 - 6x + 7

And that we can split the situation, two extensions of degree three are needed "only".

We have in a picture for $K=K_9:=\Bbb Q(c)$, the field generated by $c$, the following tower of subfields... $$ \require{AMScd} \color{darkgreen}{ \begin{CD} K=K_9=\Bbb Q(c)\\ @VV :3 V\\ K_3=\Bbb Q(g) @. \qquad\text{ where $g$ is a root of }x^3 - x^2 - 6x + 7\in\Bbb Q[x]\\ @VV :3 V\\ \Bbb Q \end{CD}} $$ And from here we have to get the possible values for $g$, then solve the corresponding equation(s) of degree three with coefficients in $K_3$ to obtain $c$. These last equations can be obtained by factorizing the minimal polynomial of $c$ over $K_3$:

sage: P = ( 2*cos(pi/19) ).minpoly()
sage: Q = x^3 - x^2 - 6*x + 7
sage: K.<g> = NumberField(Q)
sage: RK.<X> = PolynomialRing(K)
sage: factor(P(X))
    (X^3 -         g     *X^2 + ( g^2     - 5)*X - g^2     + 6)
  * (X^3 + (-g^2     + 4)*X^2 + (-g^2 - g + 4)*X + g^2 + g - 3)
  * (X^3 + ( g^2 + g - 5)*X^2 + (       g - 1)*X       - g + 2)

The factorization was manually adjusted.

This concludes the case $p=19$, one could say more, but...


The case $p=23$:

Let $u$ be the complex number $$ u = \exp \frac {\pi i}{23} = \cos \frac {\pi}{23} +i \sin \frac {\pi}{23} = \cos \frac {2\pi}{46} +i \sin \frac {2\pi}{46} \ . $$ It is a cyclotomic unit of order $46$, and satisfies the equation $u^{23}=\exp(\pi i)=-1$, so it is a root of the polynomial of degree $23-1=22=2\cdot 11$: $$ q = X^{22} - X^{21} + \dots + X^4 - X^3 + X^2 - X + 1\ . $$ The Galois group of the related cyclotomic field is cyclic and abelian, so we expect a tower of subfields corresponding to the subgroups of $\Bbb Z/22$. We expect that $$ c = 2\cos\frac \pi{23} = u + \frac 1u = u^1 + u^{22} $$ satisfies an equation of degree $22/2=11$. This is obtained as above in sage, $c$ and its conjugates are roots of the polynomial: $$ P = x^{11} - x^{10} - 10 x^{9} + 9 x^{8} + 36 x^{7} - 28 x^{6} - 56 x^{5} + 35 x^{4} + 35 x^{3} - 15 x^{2} - 6 x + 1 \ . $$ The roots of the above polynomials are $$ c_k = 2\cos\frac {k\pi}{23} = u^k + u^{23-k}\ , $$ where $k$ takes the half of the values between $1$ and $22$, we can restrict for instance to the odd ones. (We have $c_1=c_{22}$, $c_2=c_{21}$, ...)

We expect again for Galois theoretical reasons that solutions are of the shape $$ \frac 1{11} \Big(\ 1 + s_1 + s_2 + \dots s_{10} \ \Big) = \frac 1{11} \Big(\ 1 + S_1^{1/11} + S_2^{1/11} + \dots S_{10}^{1/11} \ \Big) \ , $$ with suitable, inter-coordinated, matching choices for the many radicals $\displaystyle s_k=\sqrt[11]{S_k}$ and we have an equation of degree $10$ over $\Bbb Q$ for the numbers $S_1,S_2,\dots,S_{10}$ to solve only.

This time we really need Galois theory. The number $c_1=c$ generates a field of degree $11$, a prime degree, over $\Bbb Q$. Its conjugates are $c_k$, $k=1,3,5,7,9,11,13,15,17,19$, and there is a Galois substitution $\tau$ over $\Bbb Q$ permuting them. One can take for $\tau$ the map determined by $u\to u^g$, where $g\in(\Bbb Z/23)^\times$ has order $11$, i will use $g=3$ below.

Since in such cases it is useful to have also the roots of unity of degree $11$, we will work with the following extension of fields: $$ L=\Bbb Q(c,\zeta) \text{ over } K=\Bbb Q(\zeta) \ . $$ Here, we adjoin the root of unity of degree $11$, $\zeta=\zeta_{11}$. The extension of $\tau$ as a Galois substitution of $L$, trivially action on $\zeta$ will be denoted also by $\tau$.

Then $\zeta\in K$ has norm one: $$ N_{L/K}(\zeta) =\prod_{k}\tau^k(\zeta) =\prod_{k}\zeta =1\ . $$ Hilbert 90 insures the existence of some $b\in L$ with $$ \zeta=\frac{b}{\tau(b)}\ . $$ In fact, this existence is constructive, so let us consider (the Lagrange rezolvent) $$ b = c + \zeta\tau(c) + \zeta^2\tau^2(c) + \dots + \zeta^{10}\tau^{10}(c) \ . $$ Then $$ \frac{b}{\tau(b)} = \frac{c + \zeta\tau(c) + \zeta^2\tau^2(c) + \dots + \zeta^{10}\tau^{10}(c) }{ \tau(c) + \zeta\tau^2(c) + \zeta^2\tau^3(c) + \dots + \zeta^{10}\tau^{11}(c) } =\zeta \ . $$ Then $b^{11}$ is fixed by $\tau$, because of $$ \frac{b^{11}}{\tau(b^{11})} = \frac{b^{11}}{\tau(b)^{11}} = \left(\frac{b}{\tau(b)}\right)^{11} = \zeta^{11} =1\ . $$ So it is an element of $K$, and we can proceed by building the $11$.th root of this element.

Computations in sage deliver this polynomial:

p, q = 23, 11
F.<Z> = CyclotomicField( 2*p*q )
u, z = Z^q, Z^(2*p)
g = 3

b = sum([ (u^(g^j) + 1/u^(g^j)) * z^j for j in range(q) ]) mp = (b^q).minpoly() print(f'g = {g} :: obtained minimal polynomial of degree {mp.degree()}\n{mp}')

We obtain:

g = 3 :: obtained minimal polynomial of degree 10
x^10 + 52918009*x^9 + 1085397499666778*x^8
     + 29708537267066813398222*x^7
     + 359857016915639527522755372296*x^6
     - 16252543266366411222903696822416826813*x^5
     + 342875277171018431617211044718085541208366392*x^4
     + 26970789640316423479491732431077295067790114290557038*x^3
     + 938874200966107161710606291340323350904334419758675374826774*x^2
     + 43614237139618047406820949477941119609397648573092468104280673454169*x
     + 785291652424037263548733517675617858730558099314688805476989869947753298407

(The result was broken manually.)

The free coefficient is $23^{55}$.

The coefficients in $x^9$ and $x^1$ are $23 \cdot 53 \cdot 43411$ and $23^{45} \cdot 53 \cdot 43411$.

The coefficients in $x^8$ and $x^2$ are $23^3 \cdot 2\cdot 20693 \cdot 2155519$ and $23^{36} \cdot 2\cdot 20693 \cdot 2155519$.

Similar coincidences apply for the further coefficients. From the point of view of the theory of radical extensions we are already done here. This polynomial splits in two conjugated factors if we adjoin $\sqrt{-11}$, and then we have to get "explicitly" the roots of a polynomial of degree five...

(The above polynomial splits in linear factors over $\Bbb Q(\zeta_{11})$, but the roots are not "easy", and the knowledge of an explicit expression for $\zeta_{11}$ in terms of radicals is useless.)

Things may be computed "explicitly" further, but this is not the best place to do it, and even if the job is done till the final end, i do not see a direct purpose for the obtained expressions... So let me stop here, i hope the ideas, and the computational framework can be identified and applied, when a "purpose" is present.

Oscar Lanzi
  • 39,403
dan_fulea
  • 32,856
  • I tried to do my best for the question, it is an interesting question... To get the formulas for the minimal polynomials is easy, to solve them by using only radicals is an other problem. It is possible, but making this explicit - is a problem. I know the above is too long, but there are four quite different problems solved in one answer. I do not have Mathematica, tried some commands in Wolfram Alpha, but even https://www.wolframalpha.com/input/?i=RootReduce%5BTrigToRadicals%5BCos%5BPi%2F7%5D%5D%5D did not work... I will take a look at the above link! – dan_fulea May 23 '21 at 22:06
  • Sorry for the other question. Please answer these few questions for me as I am not as advanced as you in Galois Theory. Could you please explain what $\Bbb Q(\mathrm{variable})$ means and the intuition behind it? Second question, I believe the $ζ^{\pm 1}$ denotes some type of root, if this is wrong, please tell me the intuition behind it and what this means. Thanks. I could also ask this as a question. – Тyma Gaidash May 24 '21 at 14:05
  • The main concept / framework for the computations related to roots of polynomials is the concept of a(n algebraic) number field, it is a finite extension of $\Bbb Q$. Such an extension is generated by one (or more) algebraic numbers, if $a$ is such a number, which satisfies a polynomial equation $a^n + C a^{n-1}+\dots +Sa + T$, then $\Bbb Q(a)=\Bbb Q[a]$ is the notation for the field of all rational expressions in $a$, and it turns out that each such expression is in fact a polynomial in $a$, involving only $1,a,\dots,a^{n-1}$. For instance, if $a=\sqrt[3]2$, then "only" $1,a,a^2$ are needed.. – dan_fulea May 25 '21 at 12:53
  • .. in order to write an element of $\Bbb Q[a]=\Bbb Q(a)$. The notation $\zeta$ is a typical notation for a cyclotomic root. https://en.wikipedia.org/wiki/Cyclotomic_field For instance, $\zeta_4=i$ and $\zeta_3=\cos\frac{2\pi}3+i\sin\frac{2\pi}3=\frac 12(-1+i\sqrt 3)$. The question in the OP wants an explicit form using only "radical expressions" for such cyclotomic "units" / (algebraic) numbers. Mathematica certainly has a lot of computing power for these purposes. Take a look at https://reference.wolfram.com/language/tutorial/AlgebraicNumberFields.html - It is an important field of research.. – dan_fulea May 25 '21 at 12:57
  • I see now, the “S,T,U,V” in your answer are coefficients of the a number. I see, the notation is essentially all of the numbers whose “polynomial variable” can be expressed as a factor of some coefficient and the “secondary polynomial variable”. I at least get the main idea and will look at the links. Thanks for your help. But why answer such an old question? – Тyma Gaidash May 25 '21 at 12:59
  • Try to get used with the language of number theory, many of "your problems" are well known problems, Galois theory is also doing a good wonderful job in a structural way, understanding of the theory may be a good profit for your own ideas... – dan_fulea May 25 '21 at 13:36
  • I found the question some weeks ago, and it remained open in some firefox tab... usually i answer all those opened issues, if i can ... this questions is important - i thing. As i was (very) young i had similar questions for myself, moreover, at some Olympiad like contest problem i had to understand the case of $\exp\frac{2\pi i}7$ - i did it wrong, and this was not a good ticket in my memory, years after, at the next level, Galois theory explained why i failed with "naive thoughts" (of geometrical nature)... I am also writing a booklet on math+sage, and this is a nice experimental frame... – dan_fulea May 25 '21 at 13:44
  • I will try to also add some Mathematica(TM) code in the answer, if i can manage to make it work in the alpha input line... – dan_fulea May 25 '21 at 13:45
  • With $p=11$, you use a coefficient of $467181$. This equals $3^3×11^3×13$, so you can extract factors of $11$ to be consistent with the other terms in the polynomial. – Oscar Lanzi Jul 09 '23 at 01:07
  • @dan_fulea For $p=11$, your quartic has a root $$Z_1=-\frac{11}{4} \left(89-25 \sqrt{5}-5 \sqrt{-(410+178 \sqrt{5})}\right)$$ To factor small integers is trivial, but to factor a quartic root into 2 quartic roots is not easy. Yours factors as $$-Z_1 = \frac{ab}{11}$$ $$a=\frac{11}{4}\left(-1+5\sqrt{5}+\sqrt{-10(5+\sqrt{5})}\right)$$ $$b=\frac{11}{4}\left(-31+5\sqrt{5}-\sqrt{-10(85+31\sqrt{5})}\right)$$ I've also observed this for deg-$6$ resolvents which factor into three deg-$6$ roots. And I think a root of your deg-$10$ resolvent can be factored into 5 deg-$10$ roots. Do you know why? – Tito Piezas III Jul 09 '23 at 19:12
2

Generally cosine of $\frac{\pi}{n}$ can be found in terms of cyclic infinite nested square roots of 2. without using any complex roots.

For the sake of simplicity, here I will show evaluation of simple cosine angles like $2\cos\frac{ \pi}{9}$. Others can be referred in this forum itself by referring below

Evaluating $2\cos\frac{ \pi}{9}$ using double angle formula for cosine

$2\cos\frac{ \pi}{9}$ = $\sqrt{2 + 2\cos\frac{ 2\pi}{9}}$ = $\sqrt {2 + \sqrt{2 + 2\cos\frac{ 4\pi}{9}}}$ = $\sqrt{2 + \sqrt {2 + \sqrt{2 + 2\cos\frac{ 8\pi}{9}}}}$ = $\sqrt{2 + \sqrt {2 + \sqrt{2 - 2\cos\frac{ \pi}{9}}}}$

Above cycle repeats infinitely.

Refer below for evaluating $2\cos\frac{\pi}{11}$

Representing $\cos(\frac{π}{11})$ as cyclic infinite nested square roots of $2$

This section answers solution for Evaluating $2\cos\frac{8 \pi}{25}$

You can continue to refer the underlying links in above sections itself to evaluate any $n $ in $2\cos\frac{\pi}{n}$ where n $\in Z^{+} (2\cdot n +1)$

Code to display the cyclic infinite nested square roots of 2 for $2\cos\frac{p}{q}$.

Angle is always between $\frac{1}{4}$ and $\frac{1}{2}$ radians

from IPython.display import display, Math
import time
n = int(input("Enter an odd number to get single cycle of cyclic infinite nested square root of 2\t")) # odd number forms the denominator
for i in range(n):
    if 2 ** i > n and n < 2 ** (i + 1):
        break
numerator = 2 ** (i - 2)
print("Numerator is", numerator) # Numerator for angle
halfway_of_n = (n - 1) // 2 # This determines positive or negative sign for that term

lst = [] lis2 = [] r = numerator * 2 # first operation leads to first negative sign always begin = time.time() while r != numerator: # checkpoint to complete first cycle if r > halfway_of_n: r = n - r r = r * 2 lst = lst + ['-'] else: r = r * 2 lst = lst + ['+'] lst = lst + ['+'] print("Total number of signs",len(lst)) count = lst.count('-') # Number of negative signs print('No of minus signs is/are ', count) count = lst.count('+') # Number of positive signs print('No of plus signs is/are ', count) end = time.time() #print('Time required to execute is ', end - begin, 'seconds') print(lst)

nested_sqrt = "\sqrt{2" for sign in lst[:-1]: if sign == '+': nested_sqrt += " + \sqrt{2" elif sign == '-': nested_sqrt += " - \sqrt{2" nested_sqrt += " " + lst[-1] + "..." nested_sqrt += "}" * (len(lst) - 1) # Closing curly braces for each sign except the last one nested_sqrt += "}" # Last curly brace

denominator = n # Replace with your actual denominator value

fraction = fr"2\cos\frac{{{numerator}\pi}}{{{denominator}}}"

display(Math(f"{fraction} = ")) display(Math(nested_sqrt))

**> with above method we can accurately calculate the value of cosine

angles to the required decimals**