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(here $n = 0,1,2,3,4$ so there are $10 = 2\times 5$ vertices of the first kind (lateral vertices) and 2 vertices of the second kind (axial vertices)). Note that since $\rho$ is a root of unity of odd order, $-\rho^n$ is not a root of unity, so the ten lateral vertices consist of two pentagons, laying on parallel planes, and one pentagon is rotated by an angle $\pi/5$ with respect to the second. This means that if we take, for example, the vertex $2\rho+k$, than its 5 adjacent vertices are: $\sqrt{5}k, 2\rho^2+k,2+k, -2\rho^3-k,-2\rho^4-k$.

(here $n = 0,1,2,3,4$ so there are $10 = 2\times 5$ vertices of the first kind (lateral vertices) and 2 vertices of the second kind (axial vertices)). Note that since $\rho$ is a root of unity of odd order, $-\rho^n$ is not a fifth root of unity, so the ten lateral vertices consist of two pentagons, laying on parallel planes, and one pentagon is rotated by an angle $\pi/5$ with respect to the second. This means that if we take, for example, the vertex $2\rho+k$, than its 5 adjacent vertices are: $\sqrt{5}k, 2\rho^2+k,2+k, -2\rho^3-k,-2\rho^4-k$.

(here $n = 0,1,2,3,4$ so there are $10 = 2\times 5$ vertices of the first kind and 2 vertices of the second kind).

$$(\sqrt{15})^2 = |P|^2+|Q|^2 = (4\mathbb{sin}(\frac{2\pi}{5}))^2+(2(\mathbb{sin}(\frac{2\pi}{5})-\mathbb{sin}(\frac{\pi}{5})))^2 = (4\mathbb{sin}(\frac{\pi}{5}))^2+(2(\mathbb{sin}(\frac{2\pi}{5})+\mathbb{sin}(\frac{\pi}{5})))^2$$

(here $n = 0,1,2,3,4$ so there are $10 = 2\times 5$ vertices of the first kind (lateral vertices) and 2 vertices of the second kind (axial vertices)). Note that since $\rho$ is a root of unity of odd order, $-\rho^n$ is not a root of unity, so the ten lateral vertices consist of two pentagons, laying on parallel planes, and one pentagon is rotated by an angle $\pi/5$ with respect to the second. This means that if we take, for example, the vertex $2\rho+k$, than its 5 adjacent vertices are: $\sqrt{5}k, 2\rho^2+k,2+k, -2\rho^3-k,-2\rho^4-k$.

$$(\sqrt{15})^2 = |P|^2+|Q|^2 = (4\mathbb{sin}(\frac{2\pi}{5}))^2+(2(\mathbb{sin}(\frac{2\pi}{5})-\mathbb{sin}(\frac{\pi}{5})))^2 = (4\mathbb{sin}(\frac{\pi}{5}))^2+(2(\mathbb{sin}(\frac{2\pi}{5})+\mathbb{sin}(\frac{\pi}{5})))^2$$

In a similar vein, Gauss's statements on the regular icosahedron lead to several equivalent trigonometric expressions for the edge length of the icosahedron.

One possible direction: Gauss's statements on the vertices of regular dodecahedron can be restated more explicitly: the vertices can be divided into two groups of 10 vertices, with coordinates:

$$P = \pm 4i(\mathbb{sin}(\frac{2\pi}{5}))\rho^n, Q = \pm 2i(\mathbb{sin}(\frac{2\pi}{5})-\mathbb{sin}(\frac{\pi}{5}))$$

(here the signs of $P,Q$ are not independent so there are $10=2\times 5$ vertices in this group) and:

$$P = \pm 4i(\mathbb{sin}(\frac{\pi}{5}))\rho^n, Q = \pm 2i(\mathbb{sin}(\frac{2\pi}{5})+\mathbb{sin}(\frac{\pi}{5}))$$

so the radius of circumscribing sphere is:

$$(\sqrt{15})^2 = |P|^2+|Q|^2 = (4\mathbb{sin}(\frac{2\pi}{5}))^2+(2(\mathbb{sin}(\frac{2\pi}{5})-\mathbb{sin}(\frac{\pi}{5})))^2 = (4\mathbb{sin}(\frac{\pi}{5}))^2+(2(\mathbb{sin}(\frac{2\pi}{5})+\mathbb{sin}(\frac{\pi}{5})))^2$$

I verified the correctness of these identities, and I am sure there is an easy algebraic proof of them. However, this does not mean that nothing deep is laying behind them; after all, historical identities like Diophantus's two squares identity (which can be explained by equating norms of complex numbers) and Euler's four squares identity (which can be explained by equating norms of quaternions) can be shown to be true by simply expanding the terms, but nevertheless deep and fruitfull ideas are contained in them.

So maybe these trigonometric identities can be somehow derived by Icosian algebra, and this line of thought will also lead directly to Gauss's combinations of fifth roots of unity.