This is a sketch, so there are gaps to be filled in. A similar procedure is discussed here. I am afraid I don't know an easy explanation you seek. My answer here relies on Galois theory, and I believe a similar process can be used to construct any regular $F_p$-gon if $F_p=2^{2^p}+1$ is a Fermat prime.
Let $\zeta$ denote the primitive $17$-th root of unity $$e^{\frac{2i\pi}{17}}=\cos\left(\frac{2\pi}{17}\right)+i\sin\left(\frac{2\pi}{17}\right).$$
Denote by $\Bbb K$ the extension field of $\mathbb{Q}$ generated by $\zeta$. Let $R$ be the ring $\mathbb{Z}/17\mathbb{Z}$ with the group of units $G=R^\times \cong \mathbb{Z}/16\mathbb{Z}$. Let $G_0$ be the trivial subgroup of $G$. Identify $G$ with the Galois group $\operatorname{Gal}(K/\mathbb{Q})$ via $$g\mapsto \Big(f(\zeta)\mapsto f\left(\zeta^g\right)\Big)$$ for each $g\in G$ and for each $f(x)\in \mathbb{Q}[x]$.
Since $3$ is a primitive element modulo $17$, the subgroup of $G$ generated by $3^{2^3}=3^{8}$ is a subgroup $G_1\geq G_0$ of $G$ with $[G_1:G_0]=2$. Define
$$\omega_1=\zeta^{3^0}+\zeta^{3^8}=\zeta+\zeta^{16}.$$
Then, the fixed field $\Bbb K_1$ of $G_1$ is the subfield $\Bbb K_1=\mathbb{Q}(\omega_1)$ of $\Bbb K$ which satisfies $[\Bbb K:\Bbb K_1]=2$.
Now let $G_2$ be the subgroup of $G$ generated by $3^{2^2}=3^{4}$, so that $G_2$ contains $G_1$ and $[G_2:G_1]=2$. Define
$$\omega_2=\zeta^{3^0}+\zeta^{3^4}+\zeta^{3^8}+\zeta^{3^{12}}=\zeta+\zeta^4+\zeta^{13}+\zeta^{16}.$$
Then, the fixed field $\Bbb K_2$ of $G_2$ is the subfield $\Bbb K_2=\mathbb{Q}(\omega_2)$ of $\Bbb K_1$ which satisfies $[\Bbb K_1:\Bbb K_2]=2$.
Next, let $G_3$ be the subgroup of $G_2$ generated by $3^{2^1}=3^2$, so that $G_3$ contains $G_2$ and $[G_2:G_3]=2$. Define
$$\omega_3=\zeta^{3^0}+\zeta^{3^2}+\zeta^{3^4}+\zeta^{3^6}+\zeta^{3^8}+\zeta^{3^{10}}+\zeta^{3^{12}}+\zeta^{3^{14}},$$
i.e.,
$$\omega_3=\zeta+\zeta^2+\zeta^4+\zeta^8+\zeta^{9}+\zeta^{13}+\zeta^{15}+\zeta^{16}.$$
Therefore, the fixed field $\Bbb K_3$ of $G_3$ is the subfield $\Bbb K_3=\mathbb{Q}(\omega_3)$ of $\Bbb K_2$ which satisfies $[\Bbb K_2:\Bbb K_3]=2$.
Finally, note that $[\Bbb K_3:\mathbb{Q}]=2$. Therefore, $\omega_3$ is a root of an irreducible monic quadratic polynomial in $\mathbb{Q}[x]$. Let
$$\omega_3'=\zeta^{3^1}+\zeta^{3^3}+\zeta^{3^5}+\zeta^{3^7}+\zeta^{3^9}+\zeta^{3^{11}}+\zeta^{3^{13}}+\zeta^{3^{15}},$$
so that
$$\omega_3'=\zeta^3+\zeta^5+\zeta^6+\zeta^7+\zeta^{10}+\zeta^{11}+\zeta^{12}+\zeta^{14}.$$
It can be shown that $\omega_3+\omega_3'=-1$ and $\omega_3\omega_3'=-4$. Therefore, $\omega_3$ and $\omega_3'$ are roots of the polynomial $x^2+x-4$, so
$$\Bbb K_3=\mathbb{Q}(\omega_3)\cong \mathbb{Q}[x]/(x^2+x-4),$$
and
$$\{\omega_3,\omega'_3\}=\left\{\frac{-1\pm\sqrt{17}}{2}\right\}.$$
It can be seen that
$$2\cos\left(\frac{2\pi}{17}\right)+2\cos\left(\frac{4\pi}{17}\right)+2\cos\left(\frac{8\pi}{17}\right)+2\cos\left(\frac{16\pi}{17}\right)=\omega_3=\frac{-1+\sqrt{17}}{2}.$$
Next, define $$\omega_2'=\zeta^{3^2}+\zeta^{3^6}+\zeta^{3^{10}}+\zeta^{3^{14}}$$
so that
$$\omega'_2=\zeta^2+\zeta^8+\zeta^9+\zeta^{15}.$$
Therefore, $\omega_2+\omega_2'=\omega_3$ and $\omega_2\omega_2'=-1$. This means $\omega_2$ and $\omega'_2$ are roots of the polynomial $x^2-\omega_3x-1$, so \begin{align}\Bbb K_2&=\mathbb{Q}(\omega_2)\cong \Bbb K_3[x]/(x^2-\omega_3x-1)\\&\cong \Bbb{Q}[x]/(x^4+x^3-6x^2-x+1),\end{align}
and
$$\{\omega_2,\omega'_2\}=\left\{\frac{\omega_3\pm\sqrt{\omega_3^2+4}}{2}\right\}.$$
It can be seen that
$$2\cos\left(\frac{2\pi}{17}\right)+2\cos\left(\frac{8\pi}{17}\right)=\omega_2=\textstyle\frac{\omega_3+\sqrt{\omega_3^2+4}}{2}=\frac{-1+\sqrt{17}+\sqrt{2(17-\sqrt{17})}}{4}.$$
Finally, let
$$\omega'_1=\zeta^{3^4}+\zeta^{3^{12} }=\zeta^4+\zeta^{13}.$$
Therefore, $\omega_1+\omega_1'=\omega_2$ and $\omega_1\omega_1'=\frac{\omega_2^2-\omega_2'-4}{2}=\frac{\omega_2(1+\omega_3)-\omega_3-3}{2}$. This shows that $\omega_1$ and $\omega'_1$ are roots of the polynomial $x^2-\omega_2x+\frac{\omega_2(1+\omega_3)-\omega_3-3}{2}$, so
\begin{align}\Bbb K_1&=\mathbb{Q}(\omega_1)\cong \Bbb K_2[x]/\left(x^2-\omega_2x+\frac{\omega_2(1+\omega_3)-\omega_3-3}{2}\right)
\\&\cong \Bbb K_3[x]/\Big(x^4-\omega_3x^3-(\omega_3+2)x^2+(2\omega_3+3)x-1\Big)
\\&\cong\mathbb{Q}[x]/(x^8+x^7-7x^6-6x^5+15x^4+10x^3-10x^2-4x+1),\end{align}
and
$$\{\omega_1,\omega_1'\}=\left\{\textstyle \frac{\omega_2\pm\sqrt{\omega_2^2-2\big(\omega_2(1+\omega_3)-\omega_3-3\big)}}{2}\right\}=\left\{\frac{\omega_2\pm\sqrt{2\omega_3+7-\omega_2(2+\omega_3)}}{2}\right\}.$$
It can be shown that
$$2\cos\left(\frac{2\pi}{17}\right)=\omega_1=\frac{\omega_2+\sqrt{2\omega_3+7-\omega_2(2+\omega_3)}}{2},$$
which means
$$\cos\left(\frac{2\pi}{17}\right)=\frac{-1+\sqrt{17}+\sqrt{2(17-\sqrt{17})}+2\sqrt{D}}{16},$$
where
$$D=4\big(2\omega_3+7-\omega_2(2+\omega_3)\big),$$
or
$$D=17+3\sqrt{17}-\sqrt{170+38\sqrt{17}}.$$
(Observe that $\sqrt{170+38\sqrt{17}}=\sqrt{2(17-\sqrt{17})}+2\sqrt{2(17+\sqrt{17})}$.)
By the way, you can obtain $\zeta$ by noting that
$$\zeta+\frac{1}{\zeta}=\zeta+\zeta^{16}=\omega_1.$$
Therefore, $\zeta$ (as well as $\bar\zeta=\frac{1}{\zeta}=\zeta^{16}$) is a root of the polynomial $x^2-\omega_1x+1$. That is, $\Bbb K=\mathbb{Q}(\zeta)$ satisfies
\begin{align}\Bbb K&=\mathbb{Q}(\zeta)\cong\mathbb{K}_1[x]/(x^2-\omega_1x+1)
\\&\cong\mathbb{K}_2[x]/\Big(x^4-\omega_2x^3+{\textstyle\frac{\omega_2(1+\omega_3)-\omega_3+1}{2}}x^2-\omega_2x+1\Big)
\\&\cong\mathbb{K}_3[x]/\big({\small x^8-\omega_1x^7+(2-\omega_1)x^6+(3-\omega_1)x^5+(1-2\omega_1)x^4+(3-\omega_1)x^3+(2-\omega_1)x^2-\omega_1x+1}\big)
\\&\cong\mathbb{Q}[x]/({\small x^{16}+x^{15}+x^{14}+x^{13}+x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1}).\end{align}
We have
$$\left\{\zeta,\bar{\zeta}\right\}=\left\{\frac{\omega_1\pm i\sqrt{4-\omega_1^2}}{2}\right\}.$$
Obviously,
$$\zeta=\frac{\omega_1+i\sqrt{4-\omega_1^2}}{2},$$
so that
$$\sin\left(\frac{2\pi}{17}\right)=\frac{\sqrt{4-\omega_1^2}}{2}.$$
It is too much work to write down this value, but the value of $\sin\left(\frac{2\pi}{17}\right)$ in terms of radicals can be seen here. The minimal polynomials of $\cos\left(\frac{2\pi}{17}\right)$ and $\sin\left(\frac{2\pi}{17}\right)$ in $\mathbb{Z}[x]$ are respectively.
$$\small 256x^8+128x^7-448x^6-192x^5+240x^4+80x^3-40x^2-8x+1$$
and
$$\scriptsize 65536x^{16}-278528x^{14}+487424x^{12}+452608x^{10}+239360x^8-71808x^6+11424x^4-816x^2+17.$$
I end my answer with this construction of the regular heptadecagon.
