Trigonometric functions of rational numbers are algebraic numbers!
What is the degree n of the minimal polynomial for tan ( 1 ∘ ) ?
Definition: The minimal polynomial for an algebraic number x is the monic polynomial f of degree n such that f ( x ) = 0 , f has rational coefficients, and for any f 1 such that f 1 ( x ) = 0 and f 1 has rational coefficients, de g ( f 1 ) ≥ de g ( f ) .
The answer is 24.
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2 solutions
There is a theorem that says that the degree of the minimal polynomial of tan n 2 π where n = 4 is:
- φ ( n ) if g cd ( n , 8 ) < 4
- 4 φ ( n ) if g cd ( n , 8 ) ≥ 4
Where φ is Euler's Totient function.
In this case we want the minimal polynomial of tan 3 6 0 2 π , thus n = 3 6 0 and g cd ( 3 6 0 , 8 ) = 8 , so the degree is φ ( 3 6 0 ) / 4 = 9 6 / 4 = 2 4
I don't think the theorem is correct as stated; consider n = 1 2 . You have a third case when n ≡ 4 ( m o d 8 ) ... in that case the degree of the minimal polynomial is 2 ϕ ( n ) , I believe.
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Yes, that's correct, and I wonder if there are more cases to consider.
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Here is a pretty good paper on the subject (see Page 5).There seem to be these three cases indeed.
Very nice! Do you have a proof or a reference for this theorem? If you've never seen one, there's a nice treatment (as well as a proof for the algebraic degrees of sine and cosine) in Ivan Niven's "Irrational Numbers."
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Yes, I was reading these two articles ( this and this ) about these kinds of polynomials. I am very new in this topic for now, I will read that article, thanks :D
First, it’s not too difficult to determine that the roots of the equation
x 2 − 4 x + 1 = 0
yield roots T a n ( θ ) where θ = 1 5 , 7 5 , 1 9 5 , 2 5 5
From trigonometric identities, we know that
T a n ( 3 x ) = 1 − 3 T a n ( x ) 2 3 T a n ( x ) − T a n ( x ) 3
T a n ( 5 x ) = 1 − 1 0 T a n ( x ) 2 + 5 T a n ( x ) 4 5 T a n ( x ) − 1 0 T a n ( x ) 3 + T a n ( x ) 5
so that direct consecutive substitutions would yield a polynomial of degree 3 0 .
However, 1 5 , 7 5 , 1 9 5 , 2 5 5 yields angles
5 , 2 5 , 6 5 , 8 5 , 1 2 5 , 1 4 5 , 1 8 5 , 2 0 5 , 2 4 5 , 2 6 5 , 3 0 5 , 3 2 5
which can be worked out by the formula 3 1 ( θ + 1 8 0 n ) for n = 0 , 1 , 2 , 3 … , which in turn yields angles
1 , 5 , 1 3 , 1 7 , 2 5 , 2 9 , 3 7 , 4 1 , 4 9 , 5 3 , 6 1 , 6 5 , 7 3 , 7 7 , 8 5 , 8 9 , 9 7 , 1 0 1 , 1 0 9 , 1 1 3 ,
1 2 1 , 1 2 5 , 1 3 3 , 1 3 7 , 1 4 5 , 1 4 9 , 1 5 7 , 1 6 1 , 1 6 9 , 1 7 3 , 1 8 1 , 1 8 5 , 1 9 3 , 1 9 7 , 2 0 5 ,
2 0 9 , 2 1 7 , 2 2 1 , 2 2 9 , 2 3 3 , 2 4 1 , 2 4 5 , 2 5 3 , 2 5 7 , 2 6 5 , 2 6 9 , 2 7 7 , 2 8 1 , 2 8 9 , 2 9 3 ,
3 0 1 , 3 0 5 , 3 1 3 , 3 1 7 , 3 2 5 , 3 2 9 , 3 3 7 , 3 4 1 , 3 4 9 , 3 5 3
which can be worked out by the formula 5 1 ( θ + 1 8 0 n ) for n = 0 , 1 , 2 , 3 …
Among the 6 0 angles listed, 1 2 are repeats from the previous, which means the polynomial of degree 3 0 is degenerate, a product of two polynomials of degree 6 and degree 2 4 . The revised list of angles, none of which are multiples of 3 or 5
1 , 1 3 , 1 7 , 2 9 , 3 7 , 4 1 , 4 9 , 5 3 , 6 1 , 7 3 , 7 7 , 8 9 , 9 7 , 1 0 1 , 1 0 9 , 1 1 3 , 1 2 1 , 1 3 3 ,
1 3 7 , 1 4 9 , 1 5 7 , 1 6 1 , 1 6 9 , 1 7 3 , 1 8 1 , 1 9 3 , 1 9 7 , 2 0 9 , 2 1 7 , 2 2 1 , 2 2 9 , 2 3 3 , 2 4 1 ,
2 5 3 , 2 5 7 , 2 6 9 , 2 7 7 , 2 8 1 , 2 8 9 , 2 9 3 , 3 0 1 , 3 1 3 , 3 1 7 , 3 2 9 , 3 3 7 , 3 4 1 , 3 4 9 , 3 5 3
has 4 8 entries, making for the 2 4 roots of a polynomial of degree 2 4 , since T a n ( θ ) = T a n ( θ + π )
This hardly presents a proof that the minimal polynomial for T a n ( 1 ) has degree 2 4 , but hopefully gives one an idea of why it should be so.