Why these exact values? In Mathematica I have
In[181]:= FullSimplify[ArcSin[10^(1/2)] == (Pi/2 - ArcSinh[3] I)]
Out[181]= True
In[206]:= FullSimplify[ArcSin[100^(1/3)] == (Pi/2 - ArcCosh[10^(2/3)] I)]
Out[206]= True
In[198]:= FullSimplify[ArcSin[1000^(1/4)] == (Pi/2 - ArcSinh[Sqrt[10 Sqrt[10] - 1]] I)]
Out[198]= True
In[205]:= FullSimplify[ArcSin[10000^(1/5)] == (Pi/2 - ArcCosh[10^(4/5)] I)]
Out[205]= True
If I translated what I have correctly it means that the following table of exact values is true, with emphasis on the imaginary parts.
$$
\begin{array}{| c | r |}
\hline n&                    arcsin(n) \\
\hline \\
\hline 10^{1/2}&Pi/2-arcsinh(10*.3)I \\
\hline 100^{1/3}& Pi/2-arccosh(10^{2/3})I \\
\hline 1000^{1/4}& Pi/2-arcsinh(\sqrt{10 \sqrt{10} - 1})I \\
\hline 10000^{1/5}& Pi/2-arccosh(10^{4/5})I \\
\hline 100000^{1/6}&?????????????????????
\\
\hline
\end{array}
$$
I expected some such exact values for the roots of the powers of 10 from the following question.
Help explain a new theory on small sines
But why these values?
 A: Your values for $n$ are all of the form $10^{1 - 1/k}$. One part of the table is directly explained by the addition formulae for the trigonometric functions and the relation $\cos (ix) = \cosh x$:
$$\sin \left(\frac{\pi}{2} - i\xi \right) = \sin \frac{\pi}{2}\cos (-i\xi) + \cos \frac{\pi}{2}\sin (-i\xi) = \cos(-i\xi) = \cosh \xi.$$
So, whatever value $w$ is, $$\frac{\pi}{2} - i\operatorname{Ar cosh} w$$ is one possible value of $\arcsin w$. Whether Mathematica spits out true or false depends on which branches of the arcus sinus resp Area cosinus hyperbolici it chooses.
The remaining entries (with the $\operatorname{Ar sinh}$ applications, come from the identity $$\cosh^2 x - \sinh^2 x \equiv 1,$$
so $$\sinh \xi = 3 \Rightarrow \cosh^2 \xi = 1+3^2 = 10,$$
which means that for real positive values we have $\operatorname{Ar sinh} 3 = \operatorname{Ar cosh} \sqrt{10}$, and $$\sinh \xi = \sqrt{10\sqrt{10} - 1} \Rightarrow \cosh^2 \xi = 10\sqrt{10},$$
which again for positive real values means $\operatorname{Ar sinh} \sqrt{10\sqrt{10}-1} = \operatorname{Ar cosh} 10^{3/4}$.
For $n = 10^{5/6}$, the answer would be $$\frac{\pi}{2} - i \operatorname{Ar cosh} 10^{5/6} = \frac{\pi}{2} - i \operatorname{Ar sinh} \sqrt{10^{5/3}-1}.$$
