5

Prove that $$\sin\left(\frac{\pi}{13}\right)+\sin\left(\frac{3\pi}{13}\right)+\sin\left(\frac{4\pi}{13}\right)=\frac{1}{2}\sqrt{\frac{13+3\sqrt{13}}{2}}$$

My Attempt

Let $$x = \frac{1}{2}\sqrt{\frac{13+3\sqrt{13}}{2}} \implies 16x^4-52x^2+13=0$$ And through some donkey work we can calculate the chebyshev polynomial for $\sin\left(\frac{\pi}{13}\right),\sin\left(\frac{3\pi}{13}\right),\sin\left(\frac{4\pi}{13}\right)$ which will all be the same as $\sin(n\pi)=0,\text{ for all } n \in \mathbb{I} $, so $$P(x) = 4096x^{12}-13312x^{10}+16640x^8-9984x^6+2912x^4-364x^2+13$$ where $x = \sin\left(\frac{2i\pi}{13}\right), \text{ from } 1 \le i \le 12 \text{ where } i \in \mathbb{I}$, are the roots of $P(x)$.

Now I am not getting how to connect these two into a possible solution and even it is possible (probably is), its still a pretty donkey method as you need to find the $13^{th}$ chebyshev polynomial, so if possible maybe give some another method of approach to this question.

Blue
  • 75,673
sparrow_2764
  • 869
  • 2
  • 11
  • https://math.stackexchange.com/questions/1381294/how-to-prove-an-identity-trigonometry-angles-pi-13 and https://math.stackexchange.com/a/1590338/78967 – mathlove Feb 27 '23 at 19:47

2 Answers2

3

with $ \; w = \cos \frac{2 \pi}{13} + i \sin \frac{2 \pi}{13} \; \; \; $ in mind, let $$ x = -i \left( w - w^{25} + w^3 - w^{23} + w^9 - w^{17} \right) $$

we may calculate polynomials in $x.$ We may then apply the relation $w^{26 } = 1 $ repeatedly and express the outcome as sums of $w^{25}, w^{24}, ..., w^2, w,1 $

I get $$ x^4 - 13 x^2 + 13 = w^{24} + w^{22} + \cdots w^4 + w^2 + 1 $$

with no odd exponents and all the coefficients $1$

Multiply by the (nonzero) $w^2 - 1 $ to get $ x^4 - 13 x^2 + 13 $

The original ideas were due to Gauss, see Cox chapter

Many examples were worked out by Reuschle; This comes from page 529: Reuschle 1875

enter image description here

Will Jagy
  • 139,541
  • what exactly did you do go from step 1 to step 2 ? Sorry I am not getting it... – sparrow_2764 Feb 28 '23 at 07:25
  • What do you mean by steps 1 and 2? – Will Jagy Feb 28 '23 at 17:30
  • Step (1) $x=−i(w−w^{25}+w^3−w^{23}+w^9−w^{17})$

    Step (2) $x^4−13x^2+13=w^{24}+w^{22}+⋯w^4+w^2+1$

     – sparrow_2764 Mar 01 '23 at 02:51
  • the other dude @robjohn did the transition by some matrix thingy which I only half understood because I don't know matrices and determinants but how did you do the transition – sparrow_2764 Mar 01 '23 at 02:59
  • @sparrow_2764 multiply out $x^4$ and $x^2.$ The largest exponent comes from the $\left(w^{25} \right)^4$ giving exponent 100. – Will Jagy Mar 01 '23 at 03:01
  • oh ok I finally get it so you just brute force calculate $x^{4}$ and $x^2$ and then the relation will be easy to see (probably)... Anyways thank you very much! – sparrow_2764 Mar 01 '23 at 03:04
3

Let $$ s=\sin\left(\frac{\pi}{13}\right)+\sin\left(\frac{3\pi}{13}\right)+\sin\left(\frac{4\pi}{13}\right)\tag1 $$ If $\alpha=e^{\pi i/13}$, then $$ \alpha^{13}+1=0\tag2 $$ and $$ 2is=\underbrace{\alpha+\overbrace{\quad\alpha^{12}\quad}^{-\alpha^{-1}}}_{2i\sin\left(\frac{\pi}{13}\right)}+\underbrace{\alpha^3+\overbrace{\quad\alpha^{10}\quad}^{-\alpha^{-3}}}_{2i\sin\left(\frac{3\pi}{13}\right)}+\underbrace{\alpha^4+\overbrace{\quad\alpha^9\quad}^{-\alpha^{-4}}}_{2i\sin\left(\frac{4\pi}{13}\right)}\tag3 $$ Using the approach in this answer, we get that the minimal polynomial of $x=2is$ is $$ x^5+13x^4+13x=0\tag4 $$ Since we don't want $s=0$, we have $$ x^4+13x^2+13=0\tag5 $$ which has roots $$ x\in\left\{\pm i\sqrt{\frac{13\pm3\sqrt{13}}2}\right\}\tag6 $$ Since $\frac2\pi\theta\le\sin(\theta)\le\theta$ for $0\le\theta\le\frac\pi2$, we have $\frac{16}{13}\lt s\lt\frac{8\pi}{13}$. Therefore, we must have $$ s=\sqrt{\frac{13+3\sqrt{13}}8}\tag7 $$

robjohn
  • 345,667
  • same thing here I am not getting it how you are going from step (3) to (4)... and yes I read your other answer but it is beyond my current knowledge ... I mean what is a "minimal polynomial" – sparrow_2764 Feb 28 '23 at 08:02
  • Look at the linked answer. Using the relation $\alpha^{13}+1=0$, we get a basis of $\mathbb{Z}[\alpha]$: $b=\begin{bmatrix}1&\alpha&\cdots&\alpha^{12}\end{bmatrix}^T$. Then we find the matrix, $M$, defining the action of $x=\alpha+\alpha^3+\alpha^4+\alpha^9+\alpha^{10}+\alpha^{12}$ on that basis; that is, $xb=Mb$. The characteristic polynomial of $M$ is $-\lambda(\lambda^4+13\lambda^2+13)^3$. The computational difficulty is that the matrix is $13\times13$, which means the degree of the characteristic polynomial is $13$. – robjohn Feb 28 '23 at 16:47