Evaluate cos(2π17)\cos\left(\frac{2\pi}{17} \right)

cos(2π17)=116+11617+11634217+1817+31734217234+217 \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
4 years, 4 months ago

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First, let c1,c2,c3,c8c1, c2, c3…,c8 denote Cos(2π171),Cos(2π172),Cos(2π173),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 1717 vectors from the center to the vertices of the heptadecagon. The sum of 1616 of them equals the negative of the last. Hence we have the sum of c1+c2+c3++c8=12c1+c2+c3+…+c8=-\dfrac { 1 }{ 2 } , for reason of symmetry.

Next, we make use of the Cosine identity

Cos(a+b)+Cos(ab)=2Cos(a)Cos(b)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=c1c4=12(c3+c5)c14=c1\cdot c4=\dfrac { 1 }{ 2 } \left( c3+c5 \right)
c35=c3c5=12(c2+c8)c35=c3\cdot c5=\dfrac { 1 }{ 2 } \left( c2+c8 \right)
c28=c2c8=12(c6+c7)c28=c2\cdot c8=\dfrac { 1 }{ 2 } \left( c6+c7 \right)
c67=c6c7=12(c1+c4)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+14=0c14+c35+c28+c67+\dfrac { 1 }{ 4 } =0
c14c35+c35c28+c28c67+c67c14+14=0c14\cdot c35+c35\cdot c28+c28\cdot c67+c67\cdot c14+\dfrac { 1 }{ 4 } =0
c14c28+116=0c14\cdot c28+\dfrac { 1 }{ 16 } =0
c35c67+116=0c35\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=12(c14+c28)c1428=\dfrac { 1 }{ 2 } \left( c14+c28 \right)
c3567=12(c35+c67)c3567=\dfrac { 1 }{ 2 } \left( c35+c67 \right)

Again, we can infer from all the above the following

c1428+c3567+18=0c1428+c3567+\dfrac { 1 }{ 8 } =0
c1428c3567+116=0c1428\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=116(117)c1428=\dfrac { 1 }{ 16 } \left( -1-\sqrt { 17 } \right)
c3567=116(1+17)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 c14c14 and c67c67

c14c28+116=0c14\cdot c28+\dfrac { 1 }{ 16 } =0
c14+c28=2c1428c14+c28=2c1428

c14=c1428+141+16c14282c14=c1428+\dfrac { 1 }{ 4 } \sqrt { 1+16{ c1428 }^{ 2 } }

c35c67+116=0c35\cdot c67+\dfrac { 1 }{ 16 } =0
c35+c67=2c3567c35+c67=2c3567

c67=c3567+141+16c35672c67=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 c1c1

c1c4=c14c1\cdot c4=c14
c1+c4=2c67c1+c4=2c67

c1=c67+c14+c672c1=c67+\sqrt { -c14+{ c67 }^{ 2 } }

Of course, special attention was given to signs to get us THAT one quantity c1c1, 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(2π17)=c1Cos\left( \dfrac { 2\pi }{ 17 } \right) =c1
c1=116(1+17+34217+217+31734217234+217)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,c67c14, c35, c28, c67 by a,b,c,da, 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.

Please don't ask me to compute the 257-gon.

Michael Mendrin - 4 years, 4 months ago

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Pi Han Goh - 4 years, 4 months ago

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

Rindell Mabunga - 4 years, 4 months ago

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

Pi Han Goh - 4 years, 4 months ago

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Since 1717 is a fermat prime, it is possible to obtain the exact values of sin,cos,...\sin, \cos, ... of nπ17,nZ\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 257257. :D

Satyajit Mohanty - 4 years, 4 months ago

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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.

Pi Han Goh - 4 years, 4 months ago

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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".

Michael Mendrin - 4 years, 4 months ago

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@Michael Mendrin 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.

Pi Han Goh - 4 years, 4 months ago

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@Pi Han Goh 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

2m(22n+1){ 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.

Michael Mendrin - 4 years, 4 months ago

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