Vtyjkyuk

I checked on wolframalpha.com if $e^arcsin(2)$ is transcendental, and it answers "unknown". But $e^arcsin(a)$ is trancendental for $a neq 0 $ and algebraic. I found some proofs on this result and similar ones using
Lindemann theorem ($ln a$ is transcendental if $a$ is algebraic) and some identities of inverse trig funtions and $log$ functions. I will try to put some of these proofs here.

I think most results below Wolfram unknows.

$mathbbT$ denotes the transcendetal numbers set, and $mathbbA$ the algebraic set:

$(1) : e^iarcsina in mathbbT$ for $a in mathbbT$ and is irrational for $a notin mathbbQ $.

$(2) : e^arcsina in mathbbT $ for $0 neq a in mathbbA$

$(3): arccos a in mathbbT$ for $a in mathbbA$

$(4): e^iarccos a in mathbbA $ for $a in mathbbA$

$(5) : e^arccos a in mathbbT$ for $1 neq a in mathbbA$

$(6):arctana in mathbbT$ for $0,pm i neq a in mathbbA$

$e^arctana in mathbbT$ for $0,pm i neq a in mathbbA$

$e^mathrmarcsinh: a in mathbbT$ for $a in mathbbT$ and irrational for $a notin mathbbQ$

$e^mathrmarcsinh: a in mathbbA$ for $a in mathbbA$

$e^mathrmarccosh: a in mathbbA$ for $a in mathbbA$ and transcendental for $a in mathbbT$

$mathrmarctanh: a in mathbbT$ for $0, pm 1 neq a in mathbbA$

$e^mathrmarctanh: a in mathbbT$ for $a in mathbbT$ and irrational
for $a notin mathbbQ$

$e^mathrmarctanh: a notin mathbbQ$ for $a<-1$ or $a>1$.

$e^mathrmarctanh: a in mathbbA $ for $1 neq a in mathbbA$

$(7):arcsin a_1 + cdots + arcsin a_n in mathbbT$ for $a_1, ... , a_n in mathbbA$

Some proof's:

Since $arcsin a = i ln (sqrt1-a^2 -ia)$, we can multiple by $i$ and exponentiate, which gives $ (textI) exp(i arcsin a ) = 1/(sqrt1-a^2 - ia)$, which implies $(1)$. Exponentiating $(textI)$ by $-i$ and using Gelfond theorem implies $(2)$.

(Not sure about this one). By the indentity: $(textII):arccos a = -iln(sqrt1-a^2 - ia) + pi/2$ we can arrive at a contradiction of Lindemann theorem after some manipulation. For example, in $(textII)$ multiply by $i$ and exponentiate, this results in $exp(i arccos a) = i(sqrt1-a^2 - ia)$. If $arccos a$ is algebraic, than $i(sqrt1-a^2 - ia)$ is necessarily transcendental, which is not possible since $a in mathbbA$.

Following how to prove $(1)$ we can prove $(4)$ and $(5)$.

The indentity: $arctan a = (i/2) ln((1-ia)/(1+ia))$ is enough to prove $(6)$.

I think $(7)$ can be expanded to other inverse trig functions. We can prove $(7)$ by rewritten it as $iln(sqrt1-a_1^2 - ia_1) + cdots + iln(sqrt1-a_n^2 - ia_n) = in ln(sqrt1-a_1^2 - ia_1 cdot ... cdot sqrt1-a_n^2 - ia_n)$.

Considering other indentities we can prove the rest, which is basically the same thing over and over again. Most of this is exponentiating $ln$. For example, the apparently weird result that $exp(mathrmarcsinh: a) in mathbbA$ for $a in mathbbA$, comes from the identity: $mathrmarcsinh: a = ln (sqrta^2+1 + a)$.

If the proofs are correct it would be nice if Wolfram fixed this issue. But we need to be sure about these proofs first.