# Evaluate $\cos\left(\frac{2\pi}{17} \right)$

$\cos\left(\frac{2\pi}{17} \right) \\ = \\ -\frac1{16} + \frac1{16} \sqrt{17} + \frac1{16} \sqrt{34-2\sqrt{17}} + \frac18 \sqrt{17 + 3\sqrt{17} - \sqrt{34 - 2\sqrt{17}} - 2\sqrt{34 + 2\sqrt{17} }}$

Prove that the equation above is true.

Note by Pi Han Goh
5 years, 9 months ago

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First, let $c1, c2, c3…,c8$ denote $Cos\left( \dfrac { 2\pi }{ 17 } 1 \right) , Cos\left( \dfrac { 2\pi }{ 17 } 2 \right) ,Cos\left( \dfrac { 2\pi }{ 17 } 3 \right) ,…$

Consider the $17$ vectors from the center to the vertices of the heptadecagon. The sum of $16$ of them equals the negative of the last. Hence we have the sum of $c1+c2+c3+…+c8=-\dfrac { 1 }{ 2 }$, for reason of symmetry.

Next, we make use of the Cosine identity

$Cos\left( a+b \right) +Cos\left( a-b \right) =2Cos\left( a \right) Cos\left( b \right)$

to establish the following (with sometimes a bit extra work)

$c14=c1\cdot c4=\dfrac { 1 }{ 2 } \left( c3+c5 \right)$
$c35=c3\cdot c5=\dfrac { 1 }{ 2 } \left( c2+c8 \right)$
$c28=c2\cdot c8=\dfrac { 1 }{ 2 } \left( c6+c7 \right)$
$c67=c6\cdot c7=\dfrac { 1 }{ 2 } \left( c1+c4 \right)$

From all of the above, we then can infer the following (again sometimes with a bit extra work)

$c14+c35+c28+c67+\dfrac { 1 }{ 4 } =0$
$c14\cdot c35+c35\cdot c28+c28\cdot c67+c67\cdot c14+\dfrac { 1 }{ 4 } =0$
$c14\cdot c28+\dfrac { 1 }{ 16 } =0$
$c35\cdot c67+\dfrac { 1 }{ 16 } =0$

All these things can be verified by direct computation of the trigonometric quantities. Now, let’s define a couple more variables

$c1428=\dfrac { 1 }{ 2 } \left( c14+c28 \right)$
$c3567=\dfrac { 1 }{ 2 } \left( c35+c67 \right)$

Again, we can infer from all the above the following

$c1428+c3567+\dfrac { 1 }{ 8 } =0$
$c1428\cdot c3567+\dfrac { 1 }{ 16 } =0$

Okay, now we’re finally ready to start solving some simple quadratic equations. From the two equations immediately above, we can work out

$c1428=\dfrac { 1 }{ 16 } \left( -1-\sqrt { 17 } \right)$
$c3567=\dfrac { 1 }{ 16 } \left( -1+\sqrt { 17 } \right)$

Starting to look familiar, doesn’t it? Next, solving two more such quadratic equations from the following equations which can be inferred from all the above, we can work out $c14$ and $c67$

$c14\cdot c28+\dfrac { 1 }{ 16 } =0$
$c14+c28=2c1428$

$c14=c1428+\dfrac { 1 }{ 4 } \sqrt { 1+16{ c1428 }^{ 2 } }$

$c35\cdot c67+\dfrac { 1 }{ 16 } =0$
$c35+c67=2c3567$

$c67=c3567+\dfrac { 1 }{ 4 } \sqrt { 1+16{ c3567 }^{ 2 } }$

Finally, with one more quadratic equation from the following equations from way above, we can work out $c1$

$c1\cdot c4=c14$
$c1+c4=2c67$

$c1=c67+\sqrt { -c14+{ c67 }^{ 2 } }$

Of course, special attention was given to signs to get us THAT one quantity $c1$, which I leave as an exercise to the reader. Putting it all together (with just a bit of effort in untangling one surd into two at the far right side of it), we end up with

$Cos\left( \dfrac { 2\pi }{ 17 } \right) =c1$
$c1=\dfrac { 1 }{ 16 } \left( -1+\sqrt { 17 } +\sqrt { 34-2\sqrt { 17 } } +2\sqrt { 17+3\sqrt { 17 } -\sqrt { 34-2\sqrt { 17 } } -2\sqrt { 34+2\sqrt { 17 } } } \right)$

To maybe avoid going blind from all this, it might be advisable to replace variables $c14, c35, c28, c67$ by $a, b, c, d$, etc. I’ve left the numbers in so that, at any time, the correctness of the trigonometric identities can be double-checked.

- 5 years, 9 months ago

- 5 years, 9 months ago

Since $17$ is a fermat prime, it is possible to obtain the exact values of $\sin, \cos, ...$ of $\dfrac{n\pi}{17}, n \in \mathbb Z$ in form of surds. It'll be very tedious to prove it. If somebody does it, he or she should try calculating for $257$. :D

- 5 years, 9 months ago

Because it's not a common knowledge to layman, it's better to prove that because it's a Fermat's prime, we can find its exact value like you mentioned.

I'm more concerned in finding calculating this one first before we jump into larger numbers.

- 5 years, 9 months ago

There is a straightfoward way to compute this spectacular expression in surds, but it is a bit tedious. Give me some time today, I'll try to post it here in its entirety.

I think it's a perfect example of, "This problem---when solved---becomes simple".

- 5 years, 9 months ago

Woah thanks for your response and your solution, will inspect it very thorough! Is there an easy way to prove Satyajit's claim? I looked up some Neusis construction but came up with nothing.

By the way, I will deal with your report for the moon problem some time later.

- 5 years, 9 months ago

I think Friedrich Gauss was just a kid when he proved that a 17-gon was constructible. But later on, he proved that a condition for regular polygons to be constructible is that have a number of sides of the form

${ 2 }^{ m }\prod { { (2 }^{ { 2 }^{ n } } } +1)$

so that the next non-trivial constructible polygon has 257 sides.

I'll come back to this later when I get the time. I would like to point out that in order for a value to be constructible, it has to be expressible in terms of quadratic surds. Hence anything that can be worked out through a series of quadratic equations, as I have showed with the example of the 17-gon, can be constructible. But this does put a restriction on the number of sides constructible polygons can have, namely, it's related to powers of 2. Although this was proven by Galois theory, the beginning of group theory in the 19th century, apparently when Gauss determined such a condition, he did not give a full proof of it. That came later.

In other words, proving this condition is not easy nor straightforward.

- 5 years, 9 months ago

I think I saw that equation in the construction of the 17-gon

- 5 years, 9 months ago

This is right. I saw this equation in Wikipedia but I have no idea how to prove it.

- 5 years, 9 months ago