To verify that the number $u=\cot 22.5^{\circ}$ is a root of a quadratic equation, and the number $v=\frac{1}{\sin 22.5^{\circ}}$ is a root of a quartic equation, with both equations having integer coefficients and the leading coefficient equal to 1.
The angle $22.5^{\circ}$ is a quarter of a right angle and can be constructed as follows: On the extension of the leg $C A$ of the isosceles right triangle $A B C$ from vertex $A$ outward, draw a line segment $A D=A B$, and connect points $D$ and $B$. In the isosceles triangle $D A B$, $\angle D=\frac{1}{2} \angle B A C$. This constructs a quarter of a right angle. As shown in the figure.
If the length of each leg of $\triangle A B C$ is taken as 1, then
$$
\begin{array}{l}
D A=A B-\sqrt{A C^{2}+B C^{2}}-\sqrt{2}, \\
D C=D A+A C=\sqrt{2}+1,
\end{array}
$$

and
$$
D B=\sqrt{D C^{2}+B C^{2}}=\sqrt{(\sqrt{2}+1)^{2}+1}=\sqrt{4+2 \sqrt{2}} .
$$

Thus, we have
$$
\begin{array}{l}
u=\cot 22.5^{\circ}=\frac{D C}{B C}=\sqrt{2}+1, \\
v=\frac{1}{\sin 22.5^{\circ}}=\frac{D B}{B C}=\sqrt{4+2 \sqrt{2}} .
\end{array}
$$

Therefore,
$$
(u-1)^{2}=2,
$$

Expanding and simplifying (1) and (2), we get the equations $u^{2}-2 u-1=0, v^{4}-8 v^{2}+8=0$.