Algebraic numbers that cannot be expressed using integers and elementary functions 
Can we give an explicit${^*}$ example of a real algebraic number that provably cannot be represented as an expression built from integers and elementary${^{**}}$ functions only?


${^*}$ explicit means we can write down a polynomial equation with integer coefficients having the algebraic number as a root, and an interval with rational bounds that isolates that root.
${^{**}}$ an expression built from integers and elementary function only means any valid expression in the set of elementary expressions $\mathcal{E}$ (as defined in that question at MO). Briefly, it is any finite combination of the following:


*

*the imaginary unit $i$, 

*the exponent $x\mapsto e^x$,

*the principal branch of the natural logarithm $x\mapsto\ln x$, provided $x\ne0$, and

*the multiplication function $(x,y)\mapsto x\cdot y$. 


Note that it allows to express constants $\pi$, $e$, integers, rationals, sums, powers, radicals, and also trigonometric and hyperbolic functions and their inverses, e.g. 
$$\pi=i\cdot i\cdot i\cdot \ln(i\cdot i).$$

Update: I reposted this question at MO.
 A: If I'm not mistaken, this is a profound problem and little is known about it. Some years back there was an article in the American Mathematical Monthly by Timothy Chow, What is a Closed-Form Number? (pdf here). I believe what Chow calls EL numbers are the same as the numbers you have identified. 
This area is closely connected with Schanuel's conjecture. Chow proves one conditional result that is relevant here: 


*

*If Schanuel's conjecture is true, then the algebraic numbers $\alpha$ belonging to the class EL are precisely those whose equations are solvable over $\mathbb{Q}$ (meaning the Galois group of the splitting field of the minimal polynomial for $\alpha$ is solvable). 


See corollary 1 on page 444 of Chow's paper. I'm not sure a single explicit example that answers your question is known, although I'd be absolutely delighted to be shown otherwise. 
A: This may not be what your are looking for but, after some tinkering, I found your example in fact can be expressed in radicals. Let,
$$x = 2\cos \frac{2\arctan k}{5}$$
then $x$ is a root of,
$$x^5-5x^3+5x+2\left(\frac{k^2-1}{k^2+1}\right) = 0$$
This is the DeMoivre quintic in disguise,
$$x^5+5ax^3+5a^2x+b=0$$
and is solvable in radicals. Your $\alpha$ then has the radical expression,
$$\alpha = 2\cos \frac{2\arctan 2}{5} =\left(\frac{-3-4i}{5}\right)^{1/5}+\left(\frac{-3+4i}{5}\right)^{1/5} = 1.807059\dots$$
