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For what values of $x$ is $\cos x$ transcendental? Is there any way I can figure out the values of $x$ where $\cos x$ is transcendental or do I have to check individually for every $x$ whether it is or not?

B2VSi
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2 Answers2

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I'm not an expert on this, but I'm pretty sure this is not known in general. Here are some partial results:

  • There are only countably many values of $x$ such that $\cos x$ is algebraic (because there are only countably many algebraic numbers and $\cos$ takes every value only countably many times). So $\cos x$ is "almost always" transcendental in a rather strong sense.
  • If $x\neq 0$ is algebraic, then $\cos x$ is transcendental (this follows from the Lindemann-Weierstrass theorem).
  • If $x$ is a rational multiple of $\pi$, then $\cos x$ is algebraic (this is elementary and follows from the fact that $e^{ix}$ is a root of unity). More generally, if $y$ is such that $\cos y$ is algebraic and $x/y$ is rational, then $\cos x$ is algebraic.
  • If $x$ is an algebraic irrational multiple of $\pi$, then $\cos x$ is transcendental (this follows from the Gelfond-Schneider theorem). More generally, if $y$ is such that $\cos y$ is algebraic and $x/y$ is algebraic and irrational, then $\cos x$ is transcendental.

(Note that the general transcendence theorems tend to be stated in terms of exponentials; to translate them into results about cosines, you can use the fact that $\cos x$ is algebraic iff $e^{ix}$ is algebraic.)

Eric Wofsey
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Let $T \subset (-1,1)$ be the set of transcendental numbers in the interval $(-1,1)$. Then $\cos(x)$ is transcendental for all numbers $x$ belonging to the set $\arccos(T) = \{\arccos(t): t \in T\}$. To get all the elements in the set, we can add multiple of $\pi$ to all the elements in $\arccos(T)$.

Adhvaitha
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  • @CameronWilliams This is perfectly a right answer. – Adhvaitha Dec 13 '15 at 07:20
  • @Leg Do you mean to say that if $x$ is transcendental then arccos($x$) is transcendental? – B2VSi Dec 13 '15 at 07:21
  • @CameronWilliams You question doesn't make sense. All I have said is pick a transcendental number in $t \in (-1,1)$ and look at its $\arccos(t)$. I do not talk about $\arccos(t)$ being transcendental. – Adhvaitha Dec 13 '15 at 07:23
  • @numberphile No. All the answer says is pick a transcendental number in $t \in (-1,1)$ and look at its $\arccos(t)$. I do not talk about $\arccos(t)$ being transcendental. – Adhvaitha Dec 13 '15 at 07:23
  • @CameronWilliams Pick a transcendental number, say $t$ and compute $\arccos(t)$. Do this for all transcendental numbers $t \in (-1,1)$. This forms the set $\arccos(T)$. – Adhvaitha Dec 13 '15 at 07:26
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    I guess OP wants some kind of method to determine whether $x$ belongs to the set or not. For example, does $1$ belong to the set (i.e. is $\cos 1$ transcedental)? (By the way, I'm not one of the down voters) – mickep Dec 13 '15 at 07:28
  • I feel that the set given by Leg is correct. However, it will be better if $T$ is defined in such a way in which I can determine $x$ is in the set $T$. – B2VSi Dec 13 '15 at 07:36
  • this is a trivial tautology, a useless answer. it says that the value of cos is transcendental if the input is such that the value is transcendental ... – peter Jun 10 '21 at 09:02