Trig functions with arguments a rational multiple of π are algebriac numbers. Are ALL others trancendental (almost all are)?

 

If you know the answer, please prove it or supply reference.

At least for sin and cos, no. The proof I can see goes pretty deep, into the difference between algebraic integers and algebraic numbers.

 

A number is algebraic if it is the root to a polynomial; it is an algebraic integer if it is a root for a monic polynomial. It turns out that algebraic integers are a proper subring of algebraic numbers. We can see pretty easily that for any rational number p/q, 2 cos((p/q) pi) and 2 sin((p/q) pi) are algebraic integers, as (looking only at cos for now) 2 cos((p/q) pi) = e^((p/q) i pi) + e^(- (p/q) i pi), and the latter two are algebraic integers (as they both satisfy the equation x^(2q) - 1 = 0, and that polynomial is monic). Therefore, if you take an algebraic number between 0 and 1 that is not half of an algebraic integer, it is algebraic, but can't be cosine of a rational multiple of pi. A similar proof works for sin. And you can clearly get sec and csc, as they are just inverses. I don't have an obvious proof or disproof now for tan, though.

=Uncool-

Edited May 31, 201312 yr by uncool

There are too few algebraic numbers!

 

No function can take any real number as an argument - of which transcendentals outnumber algebraic ones - and take different algebraic values for each transcendental argument.

 

You may refine the explanation a little bit by taking a subset over which the function is strictly increasing for instance.

But that isn't the question - the question is whether the functions sin(pi*x) and cos(pi*x) create a bijection between rational numbers between 0 and 1 and algebraic numbers between 0 and 1, both of which are countable.

=Uncool-

Yes, are "all others transcendental" was the original question.

 

sin(pi*x) a bijection between rational x and algebraic sin was an assertion, not the question:

Trig functions with arguments a rational multiple of π are algebriac numbers.

 

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As for sin(pi*n/d) being algebraic:

- Because sin(pi)=sin[d*(pi/d)] is rational and is a polynom of sin(pi/d), this latter is algebraic

- sin([n*(pi/d)] is a polynom of sin(pi/d) hence is algebraic.

The "all others transcendental" was the following question, as I interpreted it:

 

If cos(x) is algebraic, then is x necessarily a rational multiple of pi?

 

Coupled with the fact that cos(pi*x) is decreasing on [0, 1/2] (which is what I should have said, rather than [0, 1]) and the intermediate value theorem, that question is equivalent to the bijection question. I don't see where there was a question about the image of an uncountable set being the set of algebraic numbers.

=Uncool-

It is equivalent to the bijection question only if adding "almost all" others are transcendental, as Mathematic did properly.

I posted the question on several forums. On one of them I got an answer. Rational values for sin or cos where the denominator of the fraction (in lowest terms) is not a power of two give the answer as no. Specifically the angle for such fractions is not a rational multiple of π.

 

The proof involved using a property of Chebychev polynomials of the first kind, using the trig. definition.

 

http://en.wikipedia.org/wiki/Chebyshev_polynomials

What values do you want for sin and cos? Rational as in your last post, or algebraic like in the first?

 

By the way, cos(π/3)=0.5 is rational despite 3 not being a power of 2.

 

Could you detail what you expect from the angle and from the sine?

What values do you want for sin and cos? Rational as in your last post, or algebraic like in the first?

 

By the way, cos(π/3)=0.5 is rational despite 3 not being a power of 2.

 

Could you detail what you expect from the angle and from the sine?

You misunderstood my statement. 0.5 = 1/2 has the denominator 2. That is what I was referring to, not the angle.