How prove this $\tan{\frac{2\pi}{13}}+4\sin{\frac{6\pi}{13}}=\sqrt{13+2\sqrt{13}}$

Question

Nice Question:

show that: The follow nice trigonometry

$$\tan{\dfrac{2\pi}{13}}+4\sin{\dfrac{6\pi}{13}}=\sqrt{13+2\sqrt{13}}$$

This problem I have ugly solution, maybe someone have nice methods? Thank you

My ugly solution:

let $$A=\tan{\dfrac{2\pi}{13}}+4\sin{\dfrac{6\pi}{13}},B=\tan{\dfrac{4\pi}{13}}+4\sin{\dfrac{\pi}{13}}$$ since $$\tan{w}=2[\sin{(2w)}-\sin{(4w)}+\sin{(6w)}-\sin{(8w)}+\cdots\pm \sin{(n-1)w}]$$ where $n$ is odd,and $w=\dfrac{2k\pi}{n}$

so

$$\tan{\dfrac{2\pi}{13}}=2\left(\sin{\dfrac{4\pi}{13}}-\sin{\dfrac{5\pi}{13}}+\sin{\dfrac{\pi}{13}}+\sin{\dfrac{3\pi}{13}}-\sin{\dfrac{6\pi}{13}}+\sin{\dfrac{2\pi}{13}}\right)$$ $$\tan{\dfrac{4\pi}{13}}=2\left(\sin{\dfrac{5\pi}{13}}-\sin{\dfrac{3\pi}{13}}-\sin{\dfrac{2\pi}{13}}-\sin{\dfrac{6\pi}{13}}-\sin{\dfrac{\pi}{13}}+\sin{\dfrac{4\pi}{13}}\right)$$ then $$A^2-B^2=(A+B)(A-B)=16\left(\sin{\dfrac{\pi}{13}}+\sin{\dfrac{3\pi}{13}}+\sin{\dfrac{4\pi}{13}}\right)\left(\sin{\dfrac{2\pi}{13}}-\sin{\dfrac{5\pi}{13}}+\sin{\dfrac{6\pi}{13}}\right)=\cdots=4\sqrt{13}$$ $$AB=\cdots=6\left(\cos{\dfrac{\pi}{13}}+\cos{\dfrac{2\pi}{13}}+\cos{\dfrac{3\pi}{13}}-\cos{\dfrac{4\pi}{13}}-\cos{\dfrac{5\pi}{13}}+\cos{\dfrac{6\pi}{13}}\right)=\cdots=3\sqrt{3}$$ so $$A=\sqrt{13+2\sqrt{13}},B=\sqrt{13-2\sqrt{13}}$$

Have other nice metods?

and I know this is simlar 1982 AMM problem: How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$

But My problem is hard then AMM problem。Thank you very much!

Answer 1

The following argument is more or less a duplicate in this paper:

Let $x=e^{2\pi i/13}$. Then $$i\tan{2\pi/13}=\frac{x^2-1}{x^2+1}=\frac{x^2-x^{26}}{x^2+1}$$

(recall that $x^{13}=1$)

$$=x^2(1-x^2)(1+x^4+x^8+x^{12}+x^3+x^7)$$ $$=(x+x^2+x^5+x^6+x^9+x^{10}-x^3-x^4-x^7-x^8-x^{11}-x^{12})$$

$$4i\sin{6\pi/13}=2(x^3-x^{10})$$

So $i\tan{2\pi/13}+4i\sin{6\pi/13}=(x+x^2+x^3+x^5+x^6+x^9-x^4-x^7-x^8-x^{10}-x^{11}-x^{12})$

Recall that $1+x+x^2+\cdots+x^{12}=0$.

After some tedious computation, we arrive at

$$(x+x^2+x^3+x^5+x^6+x^9)(x^4+x^7+x^8+x^{10}+x^{11}+x^{12})$$

$$=4+x+x^3+x^4+x^9+x^{10}+x^{12}$$

The key step in the deduction is the famous exponential sum of Gauss, which gives,

$$1+2(x+x^4+x^9+x^3+x^{12}+x^{10})=\sqrt{13}.$$

Hence $$(x+x^2+x^3+x^5+x^6+x^9)(x^4+x^7+x^8+x^{10}+x^{11}+x^{12})=(7+\sqrt{13})/2$$

Recall our formula $1+x+x^2+\cdots+x^{12}=0$ again, and

$$(x+x^2+x^3+x^5+x^6+x^9-x^4-x^7-x^8-x^{10}-x^{11}-x^{12})^2=(-1)^2-4\times(7+\sqrt{13})/2$$ $$=-13-2\sqrt{13}$$

Hence $i\tan{2\pi/13}+4i\sin{6\pi/13}=\pm i\sqrt{13+2\sqrt{13}}$

and it is obvious that $\tan{2\pi/13}+4\sin{6\pi/13}=\sqrt{13+2\sqrt{13}}$, Q.E.D.

P.S. I have a strong feeling that a generalization of such an identity to all primes is possible, but I cannot work them out right now.

Answer 2

Here's an approach using some number theory. I'm no cleaner than yours, but it does apply standard techniques that might be good to know (and pretty much always work, even when there's no short solution).

Let $\zeta = \exp\bigl(\frac{2\pi i}{13}\bigr)$. Then $\zeta$ solves the 12th-order equation $p(\zeta) = \zeta^{12} + \zeta^{11} + \dots + \zeta + 1 = 0$, and no polynomial with rational coefficients of lower degree. Note that $\tan \frac{2\pi}{13} = -i\frac{\zeta - \zeta^{-1}}{\zeta + \zeta^{-1}}$, and $\sin \frac{6\pi}{13} = -i \frac{\zeta^3 - \zeta^{-3}}{2}$. Thus the problem is equivalent to showing that

$$ \xi = \frac{\zeta - \zeta^{-1}}{\zeta + \zeta^{-1}} + 2 (\zeta^3 - \zeta^{-3}) $$

solves the equation $ \xi^2 = -13 - 2\sqrt{13} $. Note that this is almost equivalent to the equation $$(\xi^2 + 13)^2 = 52$$ The difference is the choice of $\pm \sqrt{13}$ above, which can in principal be fixed with some estimations.

This latter equation must follow purely from the algebraic equation $p(\zeta) = 0$; in particular, it must hold for all other roots of $p$. This suggests that the ideas of Galois theory could help.

So, let's take some time to document the Galois theory of the algebraic integer $\zeta$, and of the field $\mathbb Q[\zeta]$. We begin by calculating its Galois group. The roots of $p$ are $\zeta,\zeta^2,\dots,\zeta^{12}$; and so the Galois group $G = \mathrm{Aut}(\mathbb Q[\zeta])$ has order $12$. Each element $f\in G$ is of the form $f_m : \zeta \mapsto \zeta^m$; noting that $f_m f_n (\zeta) = f_m( \zeta^n) = (\zeta^m)^n = \zeta^{mn}$, we see that the Galois group is abelian. Noting that $3^3 \equiv 1 \pmod {13}$, we see that $f_3$ is an automorphism of order $3$. Finally, $2^4 \equiv 3 \pmod{13}$, so $(f_2)^4 = f_3$, from which it follows that $f_2$ is an element in $G$ of order $12$. In particular, $G \cong \mathbb Z/(12)$ is cyclic, generated by (for example) $f_2$.

This means the following. An element of $\mathbb Q[\zeta]$ — i.e. a polynomial in $\zeta$ — is rational iff it it invariant under $f_2$. Every element of $\mathbb Q[\zeta]$ solves a 12th-order polynomial. Since $f_3$ and its inverse $(f_3)^{-1} = (f_3)^2 = f_9$ are the only elements of $G$ of order $3$, an element of $\mathbb Q[\zeta]$ solves a 4th-order polynomial iff it is invariant under $f_3$. Note that $f_4$ generates the subgroup of order $6$, which has index $2$ in $\mathbb Z / 6$; therefore an element solves a quadratic equation iff it is invariant under $f_4$. And so on: the subgroup of order $2$ is generated by $f_{12} : \zeta \mapsto \zeta^{12} = \zeta^{-1}$, and so elements invariant under $f_{12}$, like $\zeta + \zeta^{-1}$, solve 6th-degree polynomials.

Returning to the $\xi$ at hand, let's suppose we don't know what polynomial it's supposed to solve, and try to find it. Before continuing, let's factor a copy of $1 + z^2$ out of $p(z) - 1$, to clear denominators in $\xi$:

$$ p(z) - 1 = (z+z^{-1})(z^2 + z^3 + z^6 + z^7 + z^{10} + z^{11}) $$ $$ \frac1{\zeta + \zeta^{-1}} = -\zeta^2 - \zeta^3 - \zeta^6 - \zeta^{-6} - \zeta^{-3} - \zeta^{-2} $$ $$ \xi = \zeta^1 + \zeta^2 - \zeta^3 - \zeta^4 + \zeta^5 + \zeta^6 - \zeta^{-6} - \zeta^{-5} + \zeta^{-4} + \zeta^{-3} - \zeta^{-2} - \zeta^{-1} + 2(\zeta^3 - \zeta^{-3}) = \zeta^1 + \zeta^2 + \zeta^3 - \zeta^4 + \zeta^5 + \zeta^6 - \zeta^{-6} - \zeta^{-5} + \zeta^{-4} - \zeta^{-3} - \zeta^{-2} - \zeta^{-1}$$

Note that the orbits under $f_3$ are $\{\zeta,\zeta^3,\zeta^4\}$, $\{\zeta^2, \zeta^6,\zeta^{18} = \zeta^5\}$, and two more formed from these by $\zeta \mapsto \zeta^{-1}$. Inspection then shows that $\xi$ is in fact invariant under $f_3$, hence solves a 4th-order equation. The four roots are necessarily given by $\xi, f_2(\xi), f_4(\xi), f_8(\xi)$. Thus the equation is $(z-\xi)(z-f_2\xi)(z - f_4\xi)(z - f_8\xi)$.

Rather than multiplying this out, let's note that $f_4(\xi) = -\xi$, and $f_8(\xi) = -f_2(\xi)$. This is because $\xi$ transforms by a factor of $-1$ under the action of $f_{12} = (f_4)^3$. (Put another way, $\xi$ is pure imaginary.) Thus we're looking for the polynomial

$$ q(z) = \bigl(z^2 - \xi^2\bigr)\bigr(z^2 - f_2(\xi)^2\bigr) $$

since it will the minimal polynomial with rational coefficients solved by $\xi$.

Let us write $\alpha = \zeta + \zeta^3 + \zeta^9$, so that $\xi = \alpha + f_2(\alpha) - f_4(\alpha) - f_8(\alpha)$, and $f_2(\xi) = - \alpha + f_2(\alpha) + f_4(\alpha) - f_8(\alpha)$. Note also that the defining equation is $\alpha + f_2(\alpha) + f_4(\alpha) + f_8(\alpha) + 1 = 0$. We are reduced to calculating two numbers: $b = \xi^2 + f_2(\xi)^2$ and $c = \xi^2f_2(\xi)^2$; then $q(z) = z^4 - bz^2 + c$. We note that

$$ \alpha^2 = f_2(\alpha) + 2\zeta^4 + 2\zeta^{10} + 2\zeta^{12} = f_2(\alpha) + 2f_4(\alpha) $$ $$ \alpha \ f_4(\alpha) = (\zeta + \zeta^3 + \zeta^{-4})(\zeta^4 + \zeta^{-1} + \zeta^{-3}) = 3 + \zeta^5 + \zeta^{-2} + \zeta^{-6} + \zeta^2 + \zeta^{-5} + \zeta^6 = 3 + f_2(\alpha) + f_8(\alpha) $$

Writing $\beta = \alpha - f_4(\alpha)$, we have: $$ \xi = \beta + f_2(\beta), \quad f_2(\xi) = -\beta + f_2(\beta) $$ $$ b = \xi^2 + f_2(\xi)^2 = 2 (\beta^2 + f_2(\beta)^2) $$ $$ c = \bigl(\xi f_2(\xi)\bigr)^2 = \bigl(\beta^2 - f_2(\beta)^2\bigr)^2 $$

$$ \beta^2 = \alpha^2 + f_4(\alpha)^2 - 2\alpha \ f_4(\alpha) = f_2(\alpha) + 2f_4(\alpha) + f_8(\alpha) + 2 \alpha - 2(3 + f_2(\alpha) + f_8(\alpha)) = -6 + 2\alpha - f_2(\alpha) + 2f_4(\alpha) - f_8(\alpha)$$ $$ \beta^2 + f_2(\beta)^2 = -12 + (\alpha + f_2\alpha + f_4\alpha + f_8\alpha) = -13 $$ $$ \beta^2 - f_2(\beta)^2 = 3\alpha - 3f_2(\alpha) + 3f_4(\alpha) - 3f_8(\alpha) = 3(\alpha + f_4(\alpha) - f_2(\alpha) - f_8(\alpha)) $$

Therefore $b = -26$. Let $\gamma = \alpha + f_4\alpha$, so that $\gamma + f_2(\gamma) = -1$, and $$ \gamma \ f_2(\gamma) = (\zeta + \zeta^3 + \zeta^4 + \zeta^{-4} + \zeta^{-3} + \zeta^{-2})(\zeta^2 + \zeta^5 + \zeta^6 + \zeta^{-6} + \zeta^{-5} + \zeta^{-2}) = \zeta + \zeta^2 + \zeta^3 + \zeta^5 + 2\zeta^6 + \zeta^{-6} + 2\zeta^{-5} + 3\zeta^{-4} + 2\zeta^{-3} + 2\zeta^{-2} + 2\zeta^{-1} + (\zeta \leftrightarrow \zeta^{-1}) = 3(\zeta + \dots + \zeta^{-1}) = -3$$

The last thing to calculate is: $$ c = 9(\gamma - f_2(\gamma))^2 = 9\bigl((\gamma + f_2\gamma)^2 - 4\gamma \ f_2\gamma\bigr) = 9\bigl( 1 - 4(-3)\bigr) = 9\times 13$$

Thus $q(z) = z^4 + 26 z^2 + 9\times 13 = (z^2 + 13)^2 - 4\times 13$, completing the proof.

Answer 3

Straightforward for WA, not so by hand: Let $x = \exp(i \theta)$ be a primitive $13$th root of unity and $y = \tan(\theta) + 4 \sin(3\theta)$. Then $(x, y)$ is a common zero of the polynomials $$i (x^2-1) x^3 + 2i (x^6-1)(x^2+1) + y (x^2+1)x^3 \textrm{ and}\\ x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1.$$

The resultant of these polynomials in the variable $x$ is $(y^4-26y^2+117)^3$. The roots of this resultant are the possible values of $y$ for different choices of $x$. Since $\sin(6 \pi/13)$ is close to one the value of $y$ in this specific case must be $\sqrt{13+2\sqrt{13}}$.

Answer 4

Another way.

Firstly, we'll prove that: $$\tan\frac{2\pi}{13}=2\left(\sin\frac{2\pi}{13}+\sin\frac{4\pi}{13}-\sin\frac{6\pi}{13}-\sin\frac{8\pi}{13}+\sin\frac{10\pi}{13}+\sin\frac{12\pi}{13}\right),$$ for which it's enough to prove that: $$\sin\frac{2\pi}{13}=\sin\frac{4\pi}{13}+\sin\frac{6\pi}{13}+\sin\frac{2\pi}{13}-\sin\frac{8\pi}{13}-\sin\frac{4\pi}{13}-\sin\frac{10\pi}{13}-\sin\frac{6\pi}{13}+$$ $$+\sin\frac{12\pi}{13}+\sin\frac{8\pi}{13}-\sin\frac{12\pi}{13}+\sin\frac{10\pi}{13},$$ which is obvious.

Thus, we need to prove that: $$4\left(\sin\frac{2\pi}{13}+\sin\frac{4\pi}{13}+\sin\frac{6\pi}{13}-\sin\frac{8\pi}{13}+\sin\frac{10\pi}{13}+\sin\frac{12\pi}{13}\right)^2=13+2\sqrt{13}.$$

Now, let $\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}=x$ and $\cos\frac{4\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}=y$.

Thus, $$x+y=\frac{2\sin\frac{\pi}{13}\left(\cos\frac{2\pi}{13}+\cos\frac{4\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}\right)}{2\sin\frac{\pi}{13}}=$$ $$=\tfrac{\sin\frac{3\pi}{13}-\sin\frac{\pi}{13}+\sin\frac{5\pi}{13}-\sin\frac{3\pi}{13}+\sin\frac{7\pi}{13}-\sin\frac{5\pi}{13}+\sin\frac{9\pi}{13}-\sin\frac{7\pi}{13}+\sin\frac{11\pi}{13}-\sin\frac{9\pi}{13}+\sin\frac{13\pi}{13}-\sin\frac{11\pi}{13}}{2\sin\frac{\pi}{13}}=-\frac{1}{2}$$ and $$xy=\left(\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}\right)\left(\cos\frac{4\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}\right)=$$ $$=\frac{3}{2}\left(\cos\frac{2\pi}{13}+\cos\frac{4\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}\right)=-\frac{3}{4}.$$ Thus, $x$ and $y$ are roots of the equation: $$t^2+\frac{1}{2}t-\frac{3}{4}=0$$ and since $$x>\cos\frac{2\pi}{13}+\cos\frac{8\pi}{13}=2\cos\frac{5\pi}{13}\cos\frac{3\pi}{13}>0,$$ we obtain: $$x=\frac{\sqrt{13}-1}{4}$$ and $$y=-\frac{\sqrt{13}+1}{4}.$$ Id est, $$4\left(\sin\frac{2\pi}{13}+\sin\frac{4\pi}{13}+\sin\frac{6\pi}{13}-\sin\frac{8\pi}{13}+\sin\frac{10\pi}{13}+\sin\frac{12\pi}{13}\right)^2=$$ $$=2\left(6+\frac{1}{2}+2\left(\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}-\cos\frac{4\pi}{13}-\cos\frac{10\pi}{13}-\cos\frac{12\pi}{13}\right)\right)=$$ $$=2\left(6+\frac{1}{2}+2\cdot\frac{\sqrt{13}}{2}\right)=13+2\sqrt{13}$$ and we are done!