# Is $\mathrm{arccos}$ of an irrational algebraic number necessarily transcendental?

I know the sine of a non-zero algebraic number is necessarily transcendental; but what about the inverse cosine of an irrational algebraic number?

• @YuriyS Fixed it. I'm just referring to the arccos of an irrational number, not to the cos of the arccos, or to the arccos of the cos, or any other combination of these functions. Commented Feb 17, 2018 at 23:23
• The problem is that the title says a different thing than the text. In the text, you are asking about $\arccos$ of an algebraic number, and in the title about $\arccos$ of an irrational number. You seem to have forgotten that transcendental numbers are also irrational.
– user491874
Commented Feb 17, 2018 at 23:28
• @user8734617 Changed the text of the question just now; I think it should be clear to everyone now! ^_^ Commented Feb 17, 2018 at 23:29
• @YuriyS But isn't $\cos 1$ trascendental? Commented Feb 17, 2018 at 23:32
• Intresting at first glance.
– mick
Commented Feb 17, 2018 at 23:44

Isn't the second statement a consequence of the first?

Let $\alpha=\arccos x$, where $x$ is algebraic. However, then $y=\sqrt{1-x^2}$ is also algebraic, as a solution of the quadratic $y^2+x^2-1=0$.

Now, suppose $\alpha$ is algebraic. It follows that either $\sin\alpha=\pm\sqrt{1-x^2}=\pm y$ is transcendental - (contradiction), or that $\alpha=0$, which actually gives you one case ($\arccos 1=0$) where $\arccos$ of an algebraic number is algebraic. (However, in this case $1$ is not irrational.)

• @YuriyS I know... :'(
– user491874
Commented Feb 17, 2018 at 23:33
• @user8734617 I'm sorry, really. :( My question was about the arccos of an irrational number that ISN'T also trascendental (I've fixed the text now). Commented Feb 17, 2018 at 23:37
• @Labba So we are back to the previous version of the text (hooray!), but then you need to fix the title...
– user491874
Commented Feb 17, 2018 at 23:38
• @user8734617 Please, forgive me for all the trouble. :( So... this means that the arccos of an algebraic irrational number is in fact trascendental, right? Since you suppose that $\alpha$ is algebraic and this gets you either a contradiction or the $\arccos 1 = 0$ case... Commented Feb 17, 2018 at 23:47
• @Labba Yes, that is the idea.
– user491874
Commented Feb 17, 2018 at 23:49