In principle that goes that way:
In[2]:= ArcCos[3] // N
Out[2]= 0. + 1.76275 I
$\cos(z)=\frac{1}{2}(\exp(iz)+\exp(-iz))$ by definition with
$z=x \arccos(3)$ it follows
$2 \cos(x \arccos(3)) \sim \exp(1.76275 x)$ for big
$x$. Now one needs only
$\exp(1.76275) \equiv 3 + \sqrt{2}$ which fits seemingly
In[4]:= E^1.76275 // N
Out[4]= 5.82844
In[5]:= 3 + 2 Sqrt[2] // N
Out[5]= 5.82843
or written more careful
In[8]:= N[E^Im[ArcCos[3]], 30] - N[3 + 2 Sqrt[2], 30]
Out[8]= 0.*10^-29
this might have even a geometrical proof, because the
$x$ has gone out of it.
