Minimal polynomial for $x=\tan \left( \frac{2}{5} \arctan p \right)+\tan \left( \frac{3}{5} \arctan p \right)$ I found numerically that the minimal polynomial for:
$$x=\tan \left( \frac{2}{5}  \arctan p \right)+\tan \left( \frac{3}{5}  \arctan p \right)$$
has the following form:
$$(3p^2-1)(p^2-1)\color{blue}{x^5}-5p(p^4-10p^2+5)\color{blue}{x^4}- \\ -40p^2(3p^2-5)\color{blue}{x^3}+80p^3(p^2-7)\color{blue}{x^2}+640p^4\color{blue}{x}-256p^5=0$$
I conjecture this form for all $p \in \mathbb{R}$.

How can this result be proven? Considering, the equation is quintic in $p$ as well, the inverse function is not supposed to have a simple form.

I'm not familiar with abstract algebra or anything related. What I did was getting minimal polynomials for several values of $p$ by Wolfram Alpha and then fitting the coefficients to a polynomial in $p$ by least squares.


Another question: is the quintic equation for $p$ solvable, i.e. is there an explicit inverse function $p(x)$?

 A: Let the polynomial in the variable $x$ whose coefficients are polynomials in $p$ be denoted $q(p,x).$ Define $u=(1/5) \arctan p,$ so that the definition of $x$ in terms of $p$ is
$$x = \tan (2u)+\tan(3u).\tag{1}$$
Since $5u=\arctan p$ we have $\tan (5u)=p.$ Now we can use the addition formulas for tangent on each of $\tan 2u, \tan 3u, \tan 5u,$ and for ease of notation we write simply $t$ for $\tan u.$ When this is done we get for $x$ the expression
$$x = \frac{t^5-10t^3+5t}{3t^4-4t^2+1}.$$
And for $p$ we arrive at the expression
$$p = \frac{t(t^4-10t^2+5)}{5t^4 -10t^2+1}.$$
Note that the expression for    $x$ can take on a zero denomintor when $t=1,1,3,$ however that is irrelevant in checking that $q(p,x)=0$ is an "identity", i.e. true for all but some isolated values.
I don't have powerful enough software to verify that indeed when the above expressions for $p$ and $x$ are put into the (complicated) quintic $q(p,x)$ the result is identically zero. However I did put the whole mess in my TI90 and check it out on a lot of values of $t$ and numerically it was zero or very nearly so for those values. 
To really finish this (rather unsatisfying) proof, one would have to run the whole thing through a good CAS, maybe Wolfram could do it (I know some maple but not Wolfram, and maybe could add later the result of a maple check.) At any rate I think this approach shows one could verify the polyomial is always zero basically from trig and algebra.
Edit: Since I was eager to get this in, I failed to notice achille hui's comment outlining a procedure using resultants. [will delete if anyone wants me to]
