What is the minimum polynomial of $x = \sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6} = \cot (7.5^\circ)$? Inspired by a previous question what let  $x = \sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6} = \cot (7.5^\circ)$.  What is the minimal polynomial of $x$ ?
The theory of algebraic extensions says the degree is $4$ since we have the degree of the field extension $[\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{6}): \mathbb{Q}] = [\mathbb{Q}(\sqrt{2}, \sqrt{3}): \mathbb{Q}] =4$
Does trigonometry help us find the other three conjugate roots?


*

*$+\sqrt{2}-\sqrt{3}+\sqrt{4}-\sqrt{6} = \cot \theta_1$

*$-\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6} = \cot \theta_2$

*$-\sqrt{2}-\sqrt{3}+\sqrt{4}+\sqrt{6} = \cot \theta_3$


This problem would be easier if we used $\cos$  instead of $\cot$.  If I remember the half-angle identity or... double-angle identity:
$$ \cot \theta = \frac{\cos \theta}{\sin \theta} = \sqrt{\frac{1 - \sin \frac{\theta}{2}}{1 + \sin \frac{\theta}{2}}}$$
Sorry I am forgetting, but I am asking about the relationship between trigonometry and the Galois theory of this number.

By Galois theory the intermediate fields are $\Bbb{Q}(\sqrt2)$, $\Bbb{Q}(\sqrt3)$ and $\Bbb{Q}(\sqrt6)$. Your number is not an element of any of those, so it generates the whole 4-d extension. In particular, we know that the minimal polynomial will be a quartic. Therefore any quartic polynomial with integer coefficients with this number as a root is the minimal polynomial.
Let $x=2+\sqrt2+\sqrt3+\sqrt6$. Then
$$
0=(x-2-\sqrt3)^2-(\sqrt2+\sqrt6)^2=x^2-(4+2\sqrt3)x-1.
$$
The idea here is that squaring removes "the $\sqrt2$ content" from $\sqrt2+\sqrt6$ leaving only irrationalities coming from $\sqrt3$.
Using the obvious algebraic conjugate as an extra factor we see that
$$
m(x)=(x^2-(4+2\sqrt3)x-1)(x^2-(4-2\sqrt3)x-1)=x^4-8x^3+2x^2+8x+1
$$
fits the bill.
The algebraic conjugates of values of trig functions (at rational multiples of $\pi$) are of the same type:
$$
\begin{aligned}
2+\sqrt2+\sqrt3+\sqrt6&=\cot\frac{\pi}{24},\\
2+\sqrt2-\sqrt3-\sqrt6&=\cot\frac{17\pi}{24},\\
2-\sqrt2+\sqrt3-\sqrt6&=\cot\frac{13\pi}{24},\\
2-\sqrt2-\sqrt3+\sqrt6&=\cot\frac{5\pi}{24}.
\end{aligned}
$$
I haven't checked the details, but I'm fairly sure that when you dig for the integer multiple angle formulas for cotangent, the polynomial $m(x)$ pops out. After all, we can write $\cot 6x$ as a rational function of $\cot x$ with polynomials of degrees $\le6$ as numerators and denominators. When you use that formula in the l.h.s. of the equation
$$
\cot 6x=1
$$
and clear the denominator, the resulting degree $6$ polynomial equation in $\cot x$ factors as a product of a quartic (obviously $m(\cot x)$) and a quadratic - the latter accounting for solutions $x=3\pi/8$ and $x=7\pi/8$.
