How can one show : $$\tan\frac{2\pi}{13}\tan\frac{5\pi}{13}\tan\frac{6\pi}{13}=\sqrt{65+18\sqrt{13}}?$$

  • 1
    $\begingroup$ Incidentally,$$\sqrt{65+18\sqrt{13}}=13^{1/4}\,U_{13}^{3/2} = 13^{1/4}\Big(\tfrac{3+\sqrt{13}}2 \Big)^{3/2}$$ $\endgroup$ – Tito Piezas III Jan 26 '18 at 3:29

I delete my comments and sum them up as a possible method which can lead to a solution.

Here are my attempts:

$\textbf{Attempt 1}$. Observe that if: $A+B+C = \pi$, then $$ \tan(\pi - A) = \tan(B+C) = \frac{\tan{B}+\tan{C}}{1-\tan{B} \cdot \tan{C}}$$ So you get $$\tan{A} + \tan{B} + \tan{C} = \tan{A} \cdot \tan{B} \cdot \tan{C}$$ So now you want to find the value of $$\tan\frac{2\pi}{13} + \tan\frac{5\pi}{13} + \tan\frac{6 \pi}{13}$$ which $\mathsf{I \ don't \ know}$. You want to see this post:

As I say in the post you have to look for some equations for which $\tan$ appears as a root and then perhaps look at sum of the roots.

$\textbf{Attempt 2.}$ Consider the equation $x^{2n}-1=0$. The roots of the equation are, $$1,-1, \cos\frac{\pi}{n}+i\cdot \sin\frac{\pi}{n},\cdots, \cos\frac{(2n-1)\pi}{n}+ i \cdot \sin\frac{(2n-1)\pi}{n}$$ Therefore we can write $$ \small x^{2n}-1 = (x-1) \cdot (x+1) \cdot \biggl(x -\cos\frac{\pi}{n}-i\sin\frac{\pi}{n}\biggr) \cdots \biggl(x - \cos\frac{(n-1)\pi}{n}-i\sin\frac{(n-1)\pi}{n}\biggr)$$ Also we have $$\cos\frac{(2n-k)\pi}{n} = \cos\frac{k\pi}{n} \ \text{and} \ \sin\frac{(2n-k)\pi}{n} = -\sin\frac{k\pi}{n}$$

Using this we can pair up $$\biggl(x-\cos\frac{\pi}{n}-i\sin\frac{\pi}{n}\biggr)\cdot \biggl(x-\cos\frac{(2n-1)\pi}{n}-i\sin\frac{(2n-1)\pi}{n}\biggr)$$ $$ = x^{2} - 2\cdot x \cdot \cos\frac{\pi}{n} + 1$$

Similarly $$\biggl(x-\cos\frac{2\pi}{n}-i\sin\frac{2\pi}{n}\biggr) \cdot (x - \cos\frac{(2n-2)\pi}{n}-i \sin\frac{(2n-2)\pi}{n}$$ $$= x^{2} -2 \cdot x \cdot \cos\frac{2\pi}{n} + 1$$

Continuing in this way, we get

\begin{align*} \frac{x^{2n}-1}{x^{2}-1} &= 1 + x^{2} + x^{4} + \cdots + x^{2n-2} \\\ &= \small\biggl(x^{2} - 2x \cos\frac{\pi}{n}+1\biggr) \cdot \biggl(x^{2}-2x\cos\frac{2\pi}{n}+1\biggr) \cdots \biggl(x^{2} - 2x \cos\frac{(n-1)\pi}{n}+1\biggr) \end{align*}

Now putting $x=1$ and using the fact that $1-\cos{x} = 2 \sin^{2}\frac{x}{2}$ we get $$n = 4^{n-1} \cdot \sin^{2}\Bigl(\frac{\pi}{2n}\Bigr) \cdot \sin^{2}\Bigl(\frac{2\pi}{2n}\Bigr) \cdots \sin^{2}\Bigl(\frac{(n-1)\pi}{2n}\Bigr)$$ From this we get $$\prod\limits_{k=1}^{n-1} \sin\biggl(\frac{\pi k}{2n}\biggr) = \frac{\sqrt{n}}{2^{n-1}}$$

Following the similar procedure gives $$ \prod\limits_{k=1}^{n}\sin\biggl(\frac{k\pi}{2n+1}\biggr) = \frac{\sqrt{2n+1}}{{2^n}}, \qquad \prod\limits_{k=1}^{n}\cos\biggl(\frac{k\pi}{2n+1}\biggr) = \frac{\sqrt{n}}{{2^{n-1}}}$$

On dividing the above you get $$\prod\limits_{k=1}^{n} \tan\Bigl(\frac{k\pi}{2n+1}\Bigr) = \frac{\sqrt{2n+1}}{2^{n}} \times \frac{2^{n-1}}{\sqrt{n}} = \frac{\sqrt{2n+1}}{2 \cdot \sqrt{n}}$$


I would like to address a question nobody seems to have asked, but which should be on everybody's mind:

What's the deal with the specific values $2, 5, 6$?

Or perhaps:

How could we have anticipated that an identity like this might be true, and how might we generate others like it?

This can be explained with a little Galois theory. Let $\zeta = \exp \left( \frac{2\pi i}{13} \right)$ be a primitive $13^{th}$ root of unity and let

$$a = \frac{\zeta - \zeta^{-1}}{\zeta + \zeta^{-1}} = i \tan \frac{2\pi}{13} \in \mathbb{Q}(\zeta).$$

Since $13$ is prime, the Galois group $(\mathbb{Z}/13\mathbb{Z})^{\ast}$ of $\mathbb{Q}(\zeta)$ is cyclic of order $12$ on generator $\sigma : \zeta \mapsto \zeta^2$. This group has a unique quotient $C_4$ of order $4$, hence by the fundamental theorem of Galois theory, the unique subextension of degree $4$ is the fixed field $K$ of $\tau = \sigma^4 : \zeta \mapsto \zeta^3$. In particular,

$$z = a \tau(a) \tau^2(a) = \left( i \tan \frac{2\pi}{13} \right) \left( i \tan \frac{6 \pi}{13} \right) \left( i \tan \frac{5 \pi}{13} \right) \in K.$$

The claim is that this is equal to $-i \sqrt{ 65 + 18 \sqrt{13} }$, which also generates an extension of degree $4$. Well-known results about Gauss sums (equivalently, well-known results about ramification of primes in number fields) imply that the unique quadratic subextension of $\mathbb{Q}(\zeta)$, which is necessarily a subextension of $K$, is $F = \mathbb{Q}(\sqrt{13})$, of which $K$ is itself a quadratic extension.

This suggests that we make use of the trace and norm maps $\text{Tr}_{K/F}, N_{K/F}$. In fact, $\text{Gal}(K/F)$ is generated by the image of $\mu = \sigma^2 : \zeta \mapsto \zeta^4$, so

$$\text{Tr}_{K/F}(z) = \left( i \tan \frac{2\pi}{13} \right) \left( i \tan \frac{6 \pi}{13} \right) \left( i \tan \frac{5 \pi}{13} \right) + \left( i \tan \frac{8\pi}{13} \right) \left( i \tan \frac{11 \pi}{13} \right) \left( i \tan \frac{7 \pi}{13} \right).$$

But using the identity $\tan x = - \tan(\pi - x)$ this is clearly equal to zero. It follows that $z$ satisfies a quadratic equation of the form

$$z^2 - N_{K/F}(z) = 0$$

over $F$, so (by checking signs) it suffices to show that $N_{K/F}(z) = -65 - 18 \sqrt{13}$.

Unfortunately, I'm out of clever ideas for this last step. But at least we now know that $N_{K/F}(z) \in F$ must be of the form $a + b \sqrt{13}$, and therefore that some identity of the desired form must be true, based on a conceptual argument rather than ad hoc computations.

The above answers the question "given the LHS, why should we expect it to equal something like the RHS?" We can also ask "given the RHS, why should we expect it to equal something like the LHS?" The answer is the Kronecker-Weber theorem. It is a straightforward exercise to show that the number field generated by the RHS has Galois group $C_4$, which is in particular abelian, so by Kronecker-Weber the RHS lies in $\mathbb{Q}(\zeta_n)$ for some $n$ such that $C_4$ is a quotient of the Galois group $(\mathbb{Z}/n\mathbb{Z})^{\ast}$. Moreover, $\sqrt{13}$ must lie in $\mathbb{Q}(\zeta_n)$, and $n = 13$ is the smallest value of $n$ which satisfies both requirements. For the problem as written I think we actually need $n = 52$, or maybe $n = 26$, but you get the idea.

  • 2
    $\begingroup$ Also, $65 + 18 \sqrt(13)$ is the product of $\sqrt{13}$ by a unit in $Z[\sqrt{13}]$. Gauss sum times quadratic unit looks like it has more chance for underlying structure than some random element of the degree 2 extension. Cyclotomic units may intervene as well given that it's a product of ratios of sines and cosines of roots of unity. $\endgroup$ – zyx Aug 6 '11 at 3:59
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    $\begingroup$ Ah. Using that observation there's a short of cheap way to conclude: find a fundamental unit (I believe it's $5 + 2 \sqrt{13}$) and match the LHS to $\pm \sqrt{13}$ times the appropriate power of the fundamental unit by comparing real absolute values (say by computing the LHS to sufficient precision). $\endgroup$ – Qiaochu Yuan Aug 6 '11 at 16:41

HINT: The number $\alpha=\sqrt{65+18\sqrt{13}}$ is one of the four solutions of the equation $(\alpha^2-65)^2-4212=0$. Then, show that the LHS is also a solution of the same equation. The other three solutions must be disregarded somehow.

  • 2
    $\begingroup$ I believe you mean 4212, not 2012. $\endgroup$ – Eugene Bulkin Aug 3 '11 at 1:23
  • $\begingroup$ @Eugene Bulkin: of course, $18^2\cdot13=4212$. Thanks for pointing out the mistake, I edited my answer. $\endgroup$ – Andrea Mori Aug 3 '11 at 8:44
  • $\begingroup$ no problem. Just curious, I understand that this is obviously a solution, but how would one derive that equation from the product of tangents? $\endgroup$ – Eugene Bulkin Aug 3 '11 at 18:21
  • $\begingroup$ Well, that is certainly the more difficult point of the "exercise". In fact, I haven't tried it myself but given the knowledge of the relation to be satisfied one can take 4th powers and try to use some elementary trig identities, possibly in combination with the observation in Chandru's answer. It is clear that if this work is an ad hoc solution which uses in an essential way the prior knowledge of the degree of the relation that the number satisfies, making it not too satusfactory. $\endgroup$ – Andrea Mori Aug 3 '11 at 20:00

Here goes! This is gonna take a while!

Let $\alpha = \displaystyle\frac{\pi}{13}$, for the sake of reducing clutter; we are then searching for $\tan2\alpha\tan5\alpha\tan6\alpha$. I'm using information found here, namely the following:

  1. $\displaystyle\prod_{k=1}^6 \tan k\alpha = \sqrt{13}$
  2. $\sin\alpha+\sin3\alpha+\sin4\alpha=\sqrt{\frac{1}{8}(13+3\sqrt{13})}$ (eq. 12)
  3. $\cos^2\alpha+\cos^23\alpha+\cos^24\alpha=\frac{1}{8}(11+\sqrt{13})$ (eq. 11)

By manipulating the last one, we have $\sin^2\alpha+\sin^23\alpha+\sin^24\alpha=\frac{1}{8}(13-\sqrt{13})$.

If we let $A = \sin\alpha$, $B = \sin3\alpha$, and $C = \sin4\alpha$, then we have:

  1. $A+B+C = \sqrt{\frac{1}{8}(13+3\sqrt{13})}$, which we will denote $P$ and
  2. $A^2+B^2+C^2 = \frac{1}{8}(13-\sqrt{13})$, which we will denote $Q$.

If we let $R = ABC$ and $S = (1-A^2)(1-B^2)(1-C^2)$, then $\displaystyle\tan\alpha\tan3\alpha\tan4\alpha=\frac{R}{\sqrt{S}}$. So, the answer we will be looking for is $\frac{\sqrt{13S}}{R}$. But in order to find that, of course, we must find $R$ and $S$!

With a little work, it can be shown that $AB+AC+BC = \frac{1}{2}(P^2-Q)$. Utilizing this fact and some more expansion work, it can then be shown that $A^3+B^3+C^3=\frac{1}{2}(3PQ-P^3)+3R$. However, power reduction also shows that $A^3 = \sin^3 \alpha = \frac{3\sin\alpha-\sin3\alpha}{4}$, $B^3 = \sin^3 3\alpha = \frac{3\sin3\alpha-\sin9\alpha}{4} = \frac{3\sin3\alpha-\sin4\alpha}{4}$ and $C^3 = \sin^3 4\alpha = \frac{3\sin4\alpha-\sin12\alpha}{4}= \frac{3\sin4\alpha-\sin\alpha}{4}$, the final simplification of the latter two being made possible because $\sin(\pi-x) = \sin x$. So, $A^3+B^3+C^3 = \frac{P}{2}$ as well, and plugging that into the previous equation we found gives $R = \frac{1}{6}(P^3+P-3PQ) = \frac{1}{16}\sqrt{26-6\sqrt{13}}$. (you can simplify that last part on your own, if you're a masochist)

To find $S$, first note that $(1-A^2)(1-B^2)(1-C^2) = -(A+1)(B+1)(C+1)(A-1)(B-1)(C-1)$. $(A\pm1)(B\pm1)(C\pm1) = ABC \pm (AB+AC+BC) + (A+B+C) \pm 1 = R \pm \frac{1}{2}(P^2-Q) + P \pm 1$, so $S = -(A+1)(B+1)(C+1)(A-1)(B-1)(C-1) = \frac{1}{4}(P-Q+2)^2 - (P+R)^2 = \frac{1}{128}(11+3\sqrt{13})$.

Now we need to find $\frac{\sqrt{13S}}{R}$. With a little more work, it can be found that $\sqrt{S} = \frac{1}{16}(3+\sqrt{13})$, so:

$$\frac{\sqrt{S}}{R} = \frac{3+\sqrt{13}}{\sqrt{26-6\sqrt{13}}} = \sqrt{\frac{11+3\sqrt{13}}{13-3\sqrt{13}}}=\sqrt{\frac{260+72\sqrt{13}}{52}}=\sqrt{5+\frac{18\sqrt{13}}{13}}$$

Multiplying that last part by $\sqrt{13}$ gives us our desired $\boxed{\sqrt{65+18\sqrt{13}}}$!


It's my very old problem (from 1979). $$\sum_{k=1}^6\cos\frac{2k\pi}{13}=\frac{\sum\limits_{k=1}^62\sin\frac{\pi}{13}\cos\frac{2k\pi}{13}}{2\sin\frac{\pi}{13}}=$$ $$=\frac{\sum\limits_{k=1}^6\left(\sin\frac{(2k+1)\pi}{13}-\sin\frac{(2k-1)\pi}{13}\right)}{2\sin\frac{\pi}{13}}=\frac{\sin\frac{13\pi}{13}-\sin\frac{\pi}{13}}{2\sin\frac{\pi}{13}}=-\frac{1}{2}.$$ 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, $x+y=-\frac{1}{2}$ and $$xy=\frac{1}{2}(\cos\frac{2\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+\cos\frac{12\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{12\pi}{13}+$$ $$+\cos\frac{2\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{4\pi}{13}+\cos\frac{10\pi}{13}+\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+$$ $$+\cos\frac{4\pi}{13}+\cos\frac{12\pi}{13}+\cos\frac{2\pi}{13}+\cos\frac{8\pi}{13}+\cos\frac{4\pi}{13}+\cos\frac{6\pi}{13})=$$ $$=\frac{3}{2}\sum_{k=1}^6\cos\frac{2k\pi}{13}=-\frac{3}{4}.$$ Since $x>0$ and $y<0$, we obtain: $x=\frac{\sqrt{13}-1}{4}$ and $y=\frac{-\sqrt{13}-1}{4}$.

Now, $$\cos\frac{2\pi}{13}\cos\frac{6\pi}{13}\cos\frac{8\pi}{13}=$$ $$=\frac{1}{2}\left(\cos\frac{4\pi}{13}+\cos\frac{8\pi}{13}\right)\cos\frac{8\pi}{13}=\frac{1}{4}\left(\cos\frac{4\pi}{13}+\cos\frac{12\pi}{13}+1+\cos\frac{10\pi}{13}\right)=$$ $$=\frac{1}{4}\left(1-\frac{1+\sqrt{13}}{4}\right)=\frac{3-\sqrt{13}}{16}.$$ By another hand, $$\cos\frac{4\pi}{13}\cos\frac{10\pi}{13}+\cos\frac{4\pi}{13}\cos\frac{12\pi}{13}+\cos\frac{10\pi}{13}\cos\frac{12\pi}{13}=$$ $$=\frac{1}{2}\sum_{k=1}^6\cos\frac{2k\pi}{13}=-\frac{1}{4}$$ and $$\cos\frac{4\pi}{13}\cos\frac{10\pi}{13}\cos\frac{12\pi}{13}=$$ $$=\frac{1}{2}\left(\cos\frac{6\pi}{13}+\cos\frac{12\pi}{13}\right)\cos\frac{12\pi}{13}=\frac{1}{4}\left(\cos\frac{6\pi}{13}+\cos\frac{8\pi}{13}+1+\cos\frac{2\pi}{13}\right)=$$ $$=\frac{1}{4}\left(1+\frac{\sqrt{13}-1}{4}\right)=\frac{3+\sqrt{13}}{16}.$$ Thus, $$\sin\frac{2\pi}{13}\sin\frac{6\pi}{13}\sin\frac{8\pi}{13}=\sqrt{\frac{\left(1-\cos\frac{4\pi}{13}\right)\left(1-\cos\frac{12\pi}{13}\right)\left(1-\cos\frac{10\pi}{13}\right)}{8}}=$$ $$=\sqrt{\frac{1+\frac{1+\sqrt{13}}{4}-\frac{1}{4}-\frac{3+\sqrt{13}}{16}}{8}}=\sqrt{\frac{13+3\sqrt{13}}{128}}.$$ Id est, $$\tan\frac{2\pi}{13}\tan\frac{5\pi}{13}\tan\frac{6\pi}{13}=-\tan\frac{2\pi}{13}\tan\frac{6\pi}{13}\tan\frac{8\pi}{13}=$$ $$=-\frac{\sqrt{\frac{13+3\sqrt{13}}{128}}}{\frac{3-\sqrt{13}}{16}}=\sqrt{\frac{(13+3\sqrt{13})(3+\sqrt{13})^2}{8}}=\sqrt{65+18\sqrt{13}}.$$ Done!

  • $\begingroup$ Very nice and without using any theory of cyclotomic polynomials / extensions. +1 $\endgroup$ – Paramanand Singh Jun 3 '17 at 11:07
  • $\begingroup$ @Paramanand Singh In my opinion the Ramanujan's technique was similar. $\endgroup$ – Michael Rozenberg Jun 3 '17 at 11:14
  • $\begingroup$ Do you think a similar approach is possible for this question math.stackexchange.com/q/2306829/72031 $\endgroup$ – Paramanand Singh Jun 5 '17 at 8:38
  • $\begingroup$ @Paramanand Singh I tried. It quits very ugly. I'll try more. $\endgroup$ – Michael Rozenberg Jun 5 '17 at 9:54

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