How can we prove or disprove that: $$\forall x \in \mathbb{A}\setminus\{0\} \implies \sin{x} \in \mathbb{R}\setminus\mathbb{A}$$ Is that holds for other trig functions?
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Consider $\text{Image}(\sin x) \cap \mathbb{A} = {-1,0,1}$. Therefore, $x \in \mathbb{A}\setminus {0}\text{ and }\sin x \in \mathbb{R}\cap \mathbb{A}$ if and only if $\dfrac{\pi}{2} \in \mathbb{A}$ or $-\dfrac{\pi}{2}\in \mathbb{A}$. – SlipEternal Feb 24 '20 at 22:03
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1$\text{Image}(\sin{x}) = [-1, 1]$, no? But $\frac{1}{2} \in [-1, 1]$ and $\frac{1}{2} \in \mathbb{A}$ – Taylor Frank Feb 25 '20 at 00:34
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1Lindemann-Weierstrass Theorem. See also this – rogerl Feb 25 '20 at 00:50
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@Taylor According to this link, the only rational numbers that are algebraic integers are integers: Wikipedia – SlipEternal Feb 25 '20 at 02:28
2 Answers
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The Hermite-Lindemann (Transcendence) Theorem (which is deep) is that if $a_1,..,a_n$ are $n$ distinct algebraic numbers and $A_1,..., A_n$ are non-$0$ algebraic, then $\sum_{j=1}^nA_je^{a_j}\ne 0.$
Equivalently, if $a_1,...,a_m$ are $m$ distinct algebraics and (in case $m=1$) not all $a_j=0,$ and if $A_1,..., A_m$ are non-$0$ algebraic, then $\sum_{j=1}^mA_je^{a_j}$ is transcendental.
If $0\ne t\in A$ then with $a_1=it,\,a_2=-it,$ and $A_1=1=-A_2,$ we have $2i\sin t=A_1e^{a_1}+A_2e^{a_2}\not \in A,$ so $\sin t \not \in A.$
DanielWainfleet
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With less than Lindemann-Weierstrass: only Lindemann. If $\sin(x)\in\overline{\mathbb Q}$, $i\cos(x)$ is too. So is $e^{ix}$.
joaopa
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