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Hello to all:

I am trying to find a general formula to write exact expressions using square roots for acute angles which are integer fractions of 2, that is, with .

For example, with I'd like to find the sine and cosine of 72 degrees. Or with the sine and cosine of 360/7 degrees.

Here is a link to give you an idea what I am after. http://www.intmath.com/blog/mathematics ... ngles-6212. I would like a general formula as a function of .

Specifically, is there a way to do this using only trigonometric identities without drawing triangles like they do in http://www.cut-the-knot.org/pythagoras/cos36.shtml to get the n=5 solution? I understand once you have a "seed" that you can use trig identities to get many other angles. But it seems so far that for n=7, or n=11, or n=13 or other primes, that I have to first draw a 7, 11, 13... prime-sided regular polygon. See also http://www.cut-the-knot.org/pythagoras/ ... .shtml#use.

Thanks,

Jay

Hi Jay

Fascinating subject. I am sure you know more about this than me!

I soon found the following site which looks good. http://www.maths.surrey.ac.uk/hosted-si ... section7.3

(You are using 2pi/n whereas this site is using pi/n. [I believe .. but you need to check ....])

This gives an exact value for sin 72 deg in terms of the golden ratio.
Values for (website) n which give exact values are:
(1), 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, 48, 51, 60, 64, 68, 80, 85, 96, 102...
which all are related to Fermat's primes.

Good luck!

Ben6993 wrote:Hi Jay

Fascinating subject. I am sure you know more about this than me!

I soon found the following site which looks good. http://www.maths.surrey.ac.uk/hosted-si ... section7.3

(You are using 2pi/n whereas this site is using pi/n. [I believe .. but you need to check ....])

This gives an exact value for sin 72 deg in terms of the golden ratio.
Values for (website) n which give exact values are:
(1), 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, 48, 51, 60, 64, 68, 80, 85, 96, 102...
which all are related to Fermat's primes.

Good luck!

Hi Ben:

It looks like Gauss proved?, suggested? that you can only solve this for the Fermat primes. So it looks like n=7, 11, 13 are impossible to do? It looks like somebody solved n=9 which is 40 degrees, see http://intmstat.com/blog/2011/06/exact- ... egrees.pdf, and got a really hairy expression.

Jay

Hi Jay

Not sure if the non-Fermat's primes are disallowed or not. I doubt that they are disallowed but would need to check up.

I didn't think that n=9 looked to bad. Shorter than n=11. A lot shorter than n=44.


Why do I get the feeling of deja vu when looking at the full table in your last link? Aha, it reminds me of your binding energy formulas! You go from one n to another (like going from one nucleide to another) and seeing the same sort of formula but not quite the same. And why are the formulae so similar and yet so different? What is the simple pattern? I just hope the code of the nucleide pattern, whatever the fundamental pattern is, and you have some of the answers, is simpler to crack than the pattern for the exact fractional angles!

Ben6993 wrote:Hi Jay

Not sure if the non-Fermat's primes are disallowed or not. I doubt that they are disallowed but would need to check up.

I didn't think that n=9 looked to bad. Shorter than n=11. A lot shorter than n=44.


Why do I get the feeling of deja vu when looking at the full table in your last link? Aha, it reminds me of your binding energy formulas! You go from one n to another (like going from one nucleide to another) and seeing the same sort of formula but not quite the same. And why are the formulae so similar and yet so different? What is the simple pattern? I just hope the code of the nucleide pattern, whatever the fundamental pattern is, and you have some of the answers, is simpler to crack than the pattern for the exact fractional angles!

Looks like some people have done what Gauss could not. Do a Google search for fifth roots of unity, seventh, ninth, and eleventh. You will find some good solutions, but it is not all easy.

The one thing that is clear to me is that if you want to solve this algebraically for any root of unity, then you start by writing what you are looking for as:

(1)

Because 1 is always a root, you can rewrite this as:

(2)

Multiplying this out reproduces (1), so this proves that the remaining roots are to be found in:

(3)

As an example, the fifth root, m=5. Here:

(4)

Sort out the trivial root of 1:

(5)

You will see that this becomes:

(6)

which is the original (4). Therefore, the hard work is to solve the polynomial:

(7)

This is the general algebraic approach. If you can solve (3), than you will have all the roots for (1).

I will very shortly be posting a new paper; that will explain why I am interested in this.

Jay