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!)

## algebraic degree of sin(1 degree)

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### Re: algebraic degree of sin(1 degree)

I put sin(1 degree) in Wolfram|Alpha, found a form with only addition, multiplication and exponents, simplified it a bit, then put it back in Wolfram|Alpha to get:

281474976710656 x^48-3377699720527872 x^46+18999560927969280 x^44-66568831992070144 x^42+162828875980603392 x^40-295364007592722432 x^38+411985976135516160 x^36-452180272956309504 x^34+396366279591591936 x^32-280058255978266624 x^30+160303703377575936 x^28-74448984852135936 x^26+28011510450094080 x^24-8500299631165440 x^22+2064791072931840 x^20-397107008634880 x^18+59570604933120 x^16-6832518856704 x^14+583456329728 x^12-35782471680 x^10+1497954816 x^8-39625728 x^6+579456 x^4-3456 x^2+1

that's pretty ugly, and good luck checking it.

281474976710656 x^48-3377699720527872 x^46+18999560927969280 x^44-66568831992070144 x^42+162828875980603392 x^40-295364007592722432 x^38+411985976135516160 x^36-452180272956309504 x^34+396366279591591936 x^32-280058255978266624 x^30+160303703377575936 x^28-74448984852135936 x^26+28011510450094080 x^24-8500299631165440 x^22+2064791072931840 x^20-397107008634880 x^18+59570604933120 x^16-6832518856704 x^14+583456329728 x^12-35782471680 x^10+1497954816 x^8-39625728 x^6+579456 x^4-3456 x^2+1

that's pretty ugly, and good luck checking it.

Last edited by Demki on Tue Mar 11, 2014 10:37 pm UTC, edited 1 time in total.

### Re: algebraic degree of sin(1 degree)

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.

### Re: algebraic degree of sin(1 degree)

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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