# Algebraic numbers

1. Feb 20, 2007

### happyg1

1. The problem statement, all variables and given/known data

Prove that sin 1 (degree) is algebraic.

2. Relevant equations

"The element $$a \in K$$ is said to be algebraic of degree n over F(a field) if it satisfies a nonzero polynomial over F of degree n but no nonzero polynomial of lower degree."

3. The attempt at a solution
I thought it might work to try this formula:
$$(cos x + i sin x)^n=cos nx + i sin nx$$
then let x=sin 1
with the identity $$cos^2 x +sin^2 x =1$$
I raised this to the 90th power because then cos disappears:
$$(\sqrt{1-x^2} + ix)^{90} = sin 90=1$$

then I put this into my TI-89 because it looks scary....and IT IS SCARY.
The coefficients all are integers (VERY LARGE INTEGERS).
The part that I don't know about is if it's the smallest nonzero polynomial that sin 1 degree satisfies and does anyboby know a neater way to go about finding this ploynomial? As it is, I can't even tell if eisenstein's criterion would apply. I'd have to go through a 90th degree polynomial and try to find a prime that would work. The coefficients look like this:
-7471375560,706252528630,-41604694413840.....and there are some twice that size.
Any input would be appreciated.
CC

Last edited: Feb 20, 2007
2. Feb 20, 2007

### StatusX

If all you have to do is show its algebraic, you just need to show it satisfies some polynomial. It might not be the smallest polynomial it satisfies, but it will have this minimal polynomial as a factor.

3. Feb 21, 2007

### Gib Z

I'm not sure if your allowed to use this fact, but For any value, other than zero, that is not a rational multiply of pi, the sine, cosine, or tan of that value is transcendental and therefore algebraic. What your asking for is pi * 1/180, so is algebraic.

4. Feb 21, 2007

### dextercioby

Actually algebraic and transcendental are totally disjoint notions. "Transcendental" means "non algebraic".

And one more thing: 1/180 is a rational multiple (of "pi").

5. Feb 21, 2007

### Gib Z

Typo make Gib Z go angry!

Transcendental and therefore NOT algebraic. 1 degree is a rational multiple of pi, and therefore THE SINE of it is algebraic.

6. Feb 21, 2007

### HallsofIvy

Staff Emeritus
If what you meant to say was
That does not follow. Your statement is that if x is NOT a rational multiple of pi then sine, cosine, or tan of that value is NOT algebraic. If this is not an "if and only if" statement, it says nothing whatsoever about what happens when x IS a rational multiple of pi.

Here's my thought, although I have not worked it out. sin(30)= 1/2. You should be able to use trig identities to reduce that to a 30th degree polynomial in sin(1).

7. Feb 21, 2007

### arildno

Hmm..so there exists a rational number R so that R*"pi"=1/180?
Fascinating.

8. Feb 21, 2007

### Gib Z

No no im not saying pi/180 is rational, im saying the multiple, 1/180, is rational.

Come on guys u know what i mean >.<"

9. Feb 21, 2007

### arildno

I know, but dexter is saying that 1/180 is a rational multiple of pi.

10. Feb 21, 2007

### matt grime

Why are you using a calculator, and why do you need to find the minimal polynomial? (Those questions are supposed to make you think about why you're doing what you're doing.)

11. Feb 22, 2007

### Gib Z

Why not find the exact value of sin 1degree? Then you can model a polynomial around that value.

12. Feb 22, 2007

### matt grime

Because that won't work. That is why. Unless you have some particularly clever way in mind to evaluate sin(1) exactly that isn't just 'use trig formulae', and which would give the answer anyway, that is.

The question just asks to show sin(1) is algebraic. That is easy. Sin(30) is algebraic, and can be written as an integral polynomial expression in sin(1). C'est tout.

13. Feb 22, 2007

### happyg1

Hey,
I *thought* I had have the minimal polynomial, going by the definition that I wrote up there. My prof. agrees that it just has to be SOME polynomial. He also says that if I want to use the "sin 30 is algebriac, so then sin 1 is algebraic" I have to prove that that sin 30 is algebraic and be rigouous in my work. So I guess I need an integral polynomial for sin 30, then show that sin 1 can be written as an integral polynomial expression in sin 1. I think I need the actual polynomial.
Any thoughts?
CC

14. Feb 22, 2007

### Dick

Don't be silly. sin(30)=1/2.

15. Feb 22, 2007

### Dick

Look. If you want an explicit polynomial, let x=exp(i*(1 degree)). Then x^180=(-1). Expand x in sin(1) and cos(1) using the binomial theorem. Ignore the imaginary parts, they'll cancel.

16. Feb 22, 2007

### HallsofIvy

Staff Emeritus
I believe that is exactly what I said 7 posts back!

17. Feb 23, 2007

### happyg1

Hi,
Halls, I read what you wrote and then went to my prof and asked him, because sometimes what he expects us to do differs from the info on the forum. I was just being careful. I really want to understand this stuff but it gives me a headache sometimes. I tend to go the long way because I easily get confused. I've been struggling with this homework problem for a while now, and I'm still fuzzy on EXACTLY what I need to write down to show that this thing is algebraic. Is it just a plynomial that sin 1 satisfies? Then I can say "Since sin 1 is a root of this polynomial and the coefficients are all integers, sin 1 is an algrbraic number"?...and put it to bed?
I have a giant polynomial that works. I don't know if I should just print it out and give him that or if I should attempt to get another polynomial that's nicer. My head hurts.
CC

18. Feb 23, 2007

### matt grime

I inserted the words ''integer" and "coefficient" into the quote. The answer is: yes, that is what the definition of algebraic is.

You can either produce one explicilty (though that would be silly) or you can show one exists (this is easy).

19. Feb 23, 2007

### Dick

I would still urge you to think hard about the expression x^180 where x=exp(i*(1 degree)). It's already practically explicit. I don't think it's necessary to actually compute the coefficients - just give a recipe for getting them.

20. Feb 23, 2007

### Gib Z

Show that sin 1 satisfies a polynomial with integer coefficient, or eqivalently, rational coefficient.