algebraic degree of sin(1 degree)

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Farabor
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algebraic degree of sin(1 degree)

Postby Farabor » Tue Mar 11, 2014 9:38 pm UTC

So, I've just finished showing that sin (1 degree) is algebraic for my homework problem. I used the method of Euler's equation, with e^(i*90 degrees), which lead me to a 180 degree polynomial in sin(1 degree) (Danged overloaded term, degree!). Thus, I've got it's algebraic of degree no more than 180....my question is (Not homework, my own curiosity), is the a smaller degree polynomial that will work?

I've looked around online and seen the work to show how you can express sin(1 degree) in terms of radicals, but I couldn't immediately follow from that what minimal degree polynomial that number is the root of (since it's ugly!)


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Qaanol
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Re: algebraic degree of sin(1 degree)

Postby Qaanol » Tue Mar 11, 2014 10:33 pm UTC

wee free kings

Moole
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Re: algebraic degree of sin(1 degree)

Postby Moole » Wed Mar 12, 2014 3:08 am UTC

Since totient(180) = 48 (totient being the number of numbers coprime to 180 less than one 180 - or the order of the multiplicative group mod 180), this makes sense; the minimum polynomial is presumably the 180th cyclotomic polynomial (though I don't know how one would prove that)
Mathematical hangover (n.): The feeling one gets in the morning when they realize that that short, elementary proof of the Riemann hypothesis that they came up with at midnight the night before is, in fact, nonsense.

DavCrav
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Re: algebraic degree of sin(1 degree)

Postby DavCrav » Fri Mar 14, 2014 5:05 am UTC

Moole wrote:Since totient(180) = 48 (totient being the number of numbers coprime to 180 less than one 180 - or the order of the multiplicative group mod 180), this makes sense; the minimum polynomial is presumably the 180th cyclotomic polynomial (though I don't know how one would prove that)


If it's true that the minimal polynomial is the 180th cyclotomic polynomial, then here is a two-step proof:

1) Prove that x^180=1.
2) Prove that x^i =/=1 for any i<180 dividing 180.

I'm sure you had that already, right?

The 180th cyclotomic polynomial is [imath]q^48+q^42-q^30-q^24-q^18+q^6+1[/imath], and substituting in [imath]sin(pi/180)[/imath] into this yields (with 100dp of precision)
[math]1.000000000028257381208207266819540259222351705966041330163593813296300859105496025205046074242117523[/math]
So it doesn't look good.


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