On generalizing $\frac{17-\sqrt{17}}8 =\sin^2(t)+\sin^2(2t)+\sin^2(4t)+\sin^2(8t)$? Given $\color{blue}{t = 2\pi/p}$ for the appropriate prime $p=4m+1$.
I. Sine
$$\begin{align}
\frac{5+\sqrt{5}}8 &=\sin^2(t)\\ 
\frac{13+\sqrt{13}}8 &=\sin^2(t)+\sin^2(3t)+\sin^2(4t)\\
\frac{17+\sqrt{17}}8 &=\sin^2(3t)+\sin^2(5t)+\sin^2(6t)+\sin^2(7t)\\
\frac{29+\sqrt{29}}8 &=\sum_{k=1}^7\sin^2(a_k\, t)
\end{align}$$
with the seven $a_k = 1,4,5,6,7,9,13.$ And so on for other prime $p=4m+1.$ For the opposite sign, one uses the remaining integers $b_k \leq \frac{p-1}2.$ For example,
$$\frac{17-\sqrt{17}}8 =\sin^2(t)+\sin^2(2t)+\sin^2(4t)+\sin^2(8t)$$
where the $a_k$ are simply $2^n$. (The next Fermat prime $p=257$ isn't so nice since it has $256/4 = 64$ sine terms.)

II. Cosine
This uses the same set of multipliers $a_k$.
$$\begin{align}
\frac{3-\sqrt{5}}8 &=\cos^2(t)\\ 
\frac{11-\sqrt{13}}8 &=\cos^2(t)+\cos^2(3t)+\cos^2(4t)\\
\frac{15-\sqrt{17}}8 &=\cos^2(3t)+\cos^2(5t)+\cos^2(6t)+\cos^2(7t)\\
\frac{27-\sqrt{29}}8 &=\sum_{k=1}^7\cos^2(a_k\, t)
\end{align}$$
with the same seven $a_k = 1,4,5,6,7,9,13.$ For the opposite sign,
$$\frac{15+\sqrt{17}}8 =\cos^2(t)+\cos^2(2t)+\cos^2(4t)+\cos^2(8t)$$

III. Conclusion
Given prime $p=4m+1$ and $t = 2\pi/p.$ The pattern clearly is,
$$\frac{p\pm\sqrt{p}}8 = \sum_{k=1}^m \sin^2(a_k\, t)$$
$$\frac{(p-2)\mp\sqrt{p}}8 = \sum_{k=1}^m \cos^2(a_k\, t)$$
Question: I used Mathematica's integer relations to find the above examples. But, for any prime $p=4m+1$, what is a clever and faster algorithm to derive the correct set of $a_k$?

The ones that have all even or all odd quadratic residues relative to p, will do the trick.
The proof for this lies is polygon isomorphisms.  The numbers are invariant if taken to an e power (ie an even quadratic), but not an odd power.  So you end up with two numbers a, b whose sum would give p, and the difference is something in sqrt(p),
