How to prove that the problem cannot be solved by the four Arithmetic Operations?

The original prolbem is as in the figure:

Suppose the square has unit side length, find the area of blue region. enter image description here

The exact solution is: $$\begin{aligned}S=&\frac{\pi-\sqrt{7}}{4}+2 \arccos\left(\frac{5\sqrt{2}}{8}\right)-\frac{1}{4}\arccos\left(-\frac{3}{4}\right)\\ \approx& 0.49263564433675266\end{aligned}$$

Since the areas of the inscribed circle and the tangent sections are rational numbers times $\pi$, I guess the exact solution cannot be obtained by the four arithmetic operations between $\pi$ and rational numbers (i.e., one has to use more advanced methods, e.g., integral, cosine theorem of triangles and so on, than primary math).

The problem here therefore can be converted into:
$$2 \arccos\left(\frac{5\sqrt{2}}{8}\right)-\frac{1}{4}\arccos\left(-\frac{3}{4}\right)-\frac{\sqrt{7}}{4}$$ cannot be represented by $\pi$ and rational numbers through addition, subtraction, multiplication and division.

But how to prove it?

Solution 1:

enter image description here

In figure 1, G is the intersection of the circles $C_1: x^2 + y^2 = (2s)^2$ and $C_2: (x – s)^2 + (y – s)^2 = s^2$.

For some $k$, $C_3 : C_1 + kC_2 = 0$ is a family of circles passing through G (and G’).

For a suitable k, we can generate the circle $C_3 : (x – 2s)^2 + (y – 2s)^2 = (\sqrt (2)s)^2$, which has center at $C(2s, 2s)$ and radius $= \sqrt (2)s$. G also lies on this circle and therefore $CG = \sqrt (2)s$.

Furthermore, all of the following can then be found:-

(1) area of ⊿AGC (with sides $2s, \sqrt (2)s, 2\sqrt (2)s$);

(2) angles $GCE, GAH, GEC, GEK, GEJ, GAB, BAG$ can all be calculated; and

(3) areas of sectors $ABG$, and $GAH$.

Move on to figure 2. It shows how the purple region can be found.

enter image description here