Trigonometry of regular polgons

In school we learned that the exterior angle of a regular n-sided polygon is $360/n$ degrees or $2\pi/n$ radians.  Similarly the interior angle of a regular n-side polygon is $180(n-2)/n$ degrees or $(n-2)\pi/n$ radians.  It can be shown that the trigonometry functions (cos, sin, tan, ...) of a rational multiple of $\pi$ is an algebraic number, i.e, a root of a polynomial with integer coefficients.  Gauss showed that $n$ being a power of 2 multiplied by distinct Fermat primes (i.e. primes of the form $2^{2^m}+1$) is sufficient to ensure the polygon can be constructed with straightedge and compass and Wantzel showed that this is also necessary (http://mathworld.wolfram.com/TrigonometryAngles.html).  This implies that the expression of $\sin(k\pi/n)$, $\cos(k\pi/n)$, etc. can be written explicitly using radicals. Some expressions for these angles are given in (http://en.wikipedia.org/wiki/Exact_trigonometric_constants) for angles where the number of minimal nesting of radicals is 2 or less.

Other values of $n$ for which the minimal nesting of radicals is 2 or less include $n=40$ and $n=120$.  For instance for $n=40$, we have:

$$\sin\left(\frac{\pi}{40}\right) = \left(\frac{1-\sqrt{5}}{8}\right) \sqrt{2+\sqrt{2}} + \frac{\sqrt{2}}{8} \sqrt{2- \sqrt{2}} \sqrt{5+\sqrt{5}}$$
$$\cos\left(\frac{\pi}{40}\right) = \left(\frac{\sqrt{5}-1}{8}\right) \sqrt{2-\sqrt{2}} + \frac{\sqrt{2}}{8} \sqrt{2+\sqrt{2}} \sqrt{5+\sqrt{5}}$$
$$\sin\left(\frac{3\pi}{40}\right) = \left(\frac{-\sqrt{5}-1}{8}\right) \sqrt{2-\sqrt{2}} + \frac{\sqrt{2}}{8} \sqrt{2+ \sqrt{2}} \sqrt{5-\sqrt{5}}$$
$$\cos\left(\frac{3\pi}{40}\right) = \left(\frac{\sqrt{5}+1}{8}\right) \sqrt{2+\sqrt{2}} + \frac{\sqrt{2}}{8} \sqrt{2-\sqrt{2}} \sqrt{5-\sqrt{5}}$$
$$\sin\left(\frac{7\pi}{40}\right) = \left(\frac{\sqrt{5}+1}{8}\right) \sqrt{2+\sqrt{2}} - \frac{\sqrt{2}}{8} \sqrt{2- \sqrt{2}} \sqrt{5-\sqrt{5}}$$
$$\cos\left(\frac{7\pi}{40}\right) = \left(\frac{\sqrt{5}+1}{8}\right) \sqrt{2-\sqrt{2}} + \frac{\sqrt{2}}{8} \sqrt{2+\sqrt{2}} \sqrt{5-\sqrt{5}}$$
$$\sin\left(\frac{9\pi}{40}\right) = \left(\frac{\sqrt{5}-1}{8}\right) \sqrt{2+\sqrt{2}} + \frac{\sqrt{2}}{8} \sqrt{2- \sqrt{2}} \sqrt{5+\sqrt{5}}$$
$$\cos\left(\frac{9\pi}{40}\right) = \left(\frac{1-\sqrt{5}}{8}\right) \sqrt{2-\sqrt{2}} + \frac{\sqrt{2}}{8} \sqrt{2+\sqrt{2}} \sqrt{5+\sqrt{5}}$$

$$\sin\left(\frac{11\pi}{40}\right) = \left(\frac{1-\sqrt{5}}{8}\right) \sqrt{2-\sqrt{2}} + \frac{\sqrt{2}}{8} \sqrt{2+ \sqrt{2}} \sqrt{5+\sqrt{5}}$$
$$\cos\left(\frac{11\pi}{40}\right) = \left(\frac{\sqrt{5}-1}{8}\right) \sqrt{2+\sqrt{2}} + \frac{\sqrt{2}}{8} \sqrt{2-\sqrt{2}} \sqrt{5+\sqrt{5}}$$

$$\sin\left(\frac{13\pi}{40}\right) = \left(\frac{1+\sqrt{5}}{8}\right) \sqrt{2-\sqrt{2}} + \frac{\sqrt{2}}{8} \sqrt{2+ \sqrt{2}} \sqrt{5-\sqrt{5}}$$
$$\cos\left(\frac{13\pi}{40}\right) = \left(\frac{\sqrt{5}+1}{8}\right) \sqrt{2+\sqrt{2}} - \frac{\sqrt{2}}{8} \sqrt{2-\sqrt{2}} \sqrt{5-\sqrt{5}}$$
$$\sin\left(\frac{17\pi}{40}\right) = \left(\frac{1+\sqrt{5}}{8}\right) \sqrt{2+\sqrt{2}} + \frac{\sqrt{2}}{8} \sqrt{2+ \sqrt{2}} \sqrt{5-\sqrt{5}}$$
$$\cos\left(\frac{17\pi}{40}\right) = \left(\frac{-\sqrt{5}-1}{8}\right) \sqrt{2-\sqrt{2}} + \frac{\sqrt{2}}{8} \sqrt{2+\sqrt{2}} \sqrt{5-\sqrt{5}}$$
$$\sin\left(\frac{19\pi}{40}\right) = \left(\frac{\sqrt{5}-1}{8}\right) \sqrt{2-\sqrt{2}} + \frac{\sqrt{2}}{8} \sqrt{2+ \sqrt{2}} \sqrt{5+\sqrt{5}}$$
$$\cos\left(\frac{19\pi}{40}\right) = \left(\frac{1-\sqrt{5}}{8}\right) \sqrt{2+\sqrt{2}} + \frac{\sqrt{2}}{8} \sqrt{2-\sqrt{2}} \sqrt{5+\sqrt{5}}$$