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.

All you need to do is write your number and its square, cube and fourth powers in the form $z^j = a_j + b_j \sqrt{2} + c_j \sqrt{3} + d_j \sqrt{6}$ with $a_j,b_j,c_j,d_j$ rational and solve a system of four equations in four unknowns 
$$ 
\eqalign{a_4 + a_3 x_3 + a_2 x_2 + a_1 x_1 + a_0 x_0  &= 0\cr
b_4 + b_3 x_3 + b_2 x_2 + b_1 x_1 + b_0 x_0  &= 0\cr
c_4 + c_3 x_3 + c_2 x_2 + c_1 x_1 + c_0 x_0  &= 0\cr
d_4 + d_3 x_3 + d_2 x_2 + d_1 x_1 + d_0 x_0  &= 0\cr
}$$
to determine a polynomial $Z^4 + x_3 Z^3 + x_2 Z^2 + x_1 Z + x_0$ of which your number is a root. 
Maple says the answer is $Z^4 - 8 Z^3 + 2 Z^2 + 8 Z + 1$.
EDIT: Here is another way, maybe a bit deeper.  Your number is $z = 1+(1+\sqrt{2})(1+\sqrt{3})$.
$u = 1+\sqrt{2}$ is an eigenvalue of $A = \pmatrix{0 & 1\cr 1 & 2\cr}$, and
$v = 1+\sqrt{3}$ is an eigenvalue of $B = \pmatrix{0 & 2\cr 1 & 2\cr}$.
So $uv = (1+\sqrt{2})(1+\sqrt{3})$ is an eigenvalue of 
$$ A \otimes B = \pmatrix{0 & 0 & 0 & 2\cr 0 & 0 & 1 & 2\cr 0 & 2 & 0 & 4\cr 1 & 2 & 2 & 4}$$
which has characteristic polynomial $X^4 - 4 X^3 - 16 X^2 - 8 X + 4$.  Substitute $X = Z - 1$ to obtain $Z^4 - 8 Z^3 + 2 Z^2 + 8 Z + 1$. 
