How does $\cos(2\pi/257)$ look like in real radicals?

Finally! Persistence pays off. This is based on Ken Brakke’s solution. However, I managed to simplify it a bit like using only 6 numbers of deg-32 (the $v_i$ below), while Brakke’s used 7 ($a_{39}, a_{40}, a_{46}, a_{47}, a_{48}, a_{54}, a_{55}$ in his site), and also simplifying the $u_i$ and $x_i$. Thus,

$$\begin{aligned} w_1&=4\cos\Big(\frac{4\,\pi}{257}\Big) = u_1+\sqrt{u_1^2-4u_2} = 3.995219\dots\\ w_2&=4\cos\Big(\frac{16\,\pi}{257}\Big) = u_3+\sqrt{u_3^2-4u_4} = 3.923736\dots\\ w_2&=4\cos\Big(\frac{64\,\pi}{257}\Big) = u_1-\sqrt{u_1^2-4u_2} = 2.837057\dots\\ w_4&=4\cos\Big(\frac{256\,\pi}{257}\Big) = u_3-\sqrt{u_3^2-4u_4} = -3.999701\dots\\ \text{where,}\\ 2u_1,\,2u_3&=v_1\pm\sqrt{v_1^2-4(v_2+v_8)}\\ 2u_2,\,2u_4&=v_9\pm\sqrt{v_9^2-4(v_{10}+v_0)}\\ \text{and,}\\ 2v_0&=x_0-\sqrt{x_0^2-4 (x_0 + x_1 + x_2 + x_5)}\\ 2v_1&=x_1-\sqrt{x_1^2-4 (x_1 + x_2 + x_3 + x_6)}\\ 2v_2&=x_2-\sqrt{x_2^2-4 (x_2 + x_3 + x_4 + x_7)}\\ 2v_8&=x_8-\sqrt{x_8^2-4 (x_8 + x_9 + x_{10} + x_{13})}\\ 2v_9&=x_9-\sqrt{x_9^2-4 (x_9 + x_{10} + x_{11} + x_{14})}\\ 2v_{10}&=x_{10}-\sqrt{x_{10}^2-4 (x_{10} + x_{11} + x_{12}+x_{15})}\\ \text{and,}\\ 2x_1,\,2x_{9}&=y_1\pm\sqrt{y_1^2-4(t_1 + y_1 + y_3 + 2 y_6)}\\ 2x_2,\,2x_{10}&=y_2\pm\sqrt{y_2^2-4(t_2 + y_2 + y_4 + 2 y_7)}\\ 2x_3,\,2x_{11}&=y_3\pm\sqrt{y_3^2-4(t_1 + y_3 + y_5 + 2 y_8)}\\ 2x_4,\,2x_{12}&=y_4\pm\sqrt{y_4^2-4(t_2 + y_4 + y_6 + 2 y_1)}\\ 2x_5,\,2x_{13}&=y_5\pm\sqrt{y_5^2-4(t_1 + y_5 + y_7 + 2 y_2)}\\ 2x_6,\,2x_{14}&=y_6\pm\sqrt{y_6^2-4(t_2 + y_6 + y_8 + 2 y_3)}\\ 2x_7,\,2x_{15}&=y_7\color{blue}{\mp}\sqrt{y_7^2-4(t_1 + y_7 + y_1 + 2 y_4)}\\ 2x_8,\,2x_{0}&=y_8\pm\sqrt{y_8^2-4(t_2 + y_8 + y_2 + 2 y_5)}\\ \text{and,}\\ 2y_1,\,2y_5&=z_1\pm\sqrt{z_1^2+4(5 +t_1 +2 z_1)}\\ 2y_2,\,2y_6&=z_2\color{blue}{\mp}\sqrt{z_2^2+4(5 +t_2 +2 z_2)}\\ 2y_3,\,2y_7&=z_3\pm\sqrt{z_3^2+4(5 +t_1 +2 z_3)}\\ 2y_4,\,2y_8&=z_4\color{blue}{\mp}\sqrt{z_4^2+4(5 +t_2 +2 z_4)}\\ \text{and,}\\ 2z_1,\,2z_3&=t_1\pm\sqrt{t_1^2+64}\\ 2z_2,\,2z_4&=t_2\pm\sqrt{t_2^2+64}\\ \text{and,}\\ t_1,\,t_2&=\frac{-1\pm\sqrt{257}}{2}\\ \end{aligned}$$

Whew! The $w_i, u_i, v_i, x_i, y_i, z_i, t_i$ of course are algebraic numbers of deg $2^7, 2^6, 2^5, 2^4, 2^3, 2^2, 2$, respectively. One can see some patterns, like how the 16 $x_i$ are so orderly expressed by the 8 $y_i$. This solution uses 24 quadratic equations $(1+2+6+8+4+2+1 = 24)$, while the one by W. Bishop involves $25$. I do not know if it can be reduced even further.

The relevant quadratic polynomials are given explicitly by Wayne Bishop, How to construct a regular polygon, American Math Monthly, March 1978, pp 186-188, available at http://poncelet.math.nthu.edu.tw/disk5/js/geometry/bishop.pdf

Although a solution has been given and more than six years have passed, I will give a solution that is a little different from the desired. I used analytical methods, but for the above reasons I will only give the final result.

$2 \cos \left (\frac{2 \pi}{2^8 +1} \right ) =$

$\sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2- \sqrt{2+ 2 \cos \left (\frac{2 \pi}{2^8 +1} \right )}}}}}}}}$