Can $\sin(\pi/25)$ be expressed in radicals I suspect that sin(pi/25) is not expressible in elementary forms in radicals because it is the root of some quintic (or rather cos(pi/25) is). Can anyone prove that that particular quintic has no solution in radicals?
I know that (from chebyshev polynomials) if $$u=\cos{\frac{\pi}{25}}$$
then
 $$u\left(16u^4-20u^2+5\right)=\frac{\phi}{2}$$ where $\phi$ is the golden ratio.

As requested, R. Israel's Maple answer can be simplified. That complicated expression,
$$u = -\frac{-3125\sqrt{-10}+12500\sqrt{5}\sqrt{\frac{\sqrt{5}}{\sqrt{5}-1}}+15625\sqrt{-2}-25000\sqrt{\frac{\sqrt{5}}{\sqrt{5}-1}}}{2048(\sqrt{5}-1)^2\sqrt{\frac{\sqrt{5}}{\sqrt{5}-1}}}$$
is just equal to,
$$u=\big(\tfrac{5}{4}\big)^5\,\Big(\tfrac{1-\sqrt{5}}{4}-\sqrt{\tfrac{5+\sqrt{5}}{8}}i\Big)$$
Hence, after some simplification, his answer reduces to the ratio of arithmetic mean (AM) and geometric mean(GM),
$$\sin\big(\tfrac{\pi}{25}\big) = \frac{AM(a,b)}{GM(a,b)} = \frac{(a+b)/2}{\sqrt{ab}} = 0.1253332\dots$$
where,
$$a = e^{4\pi i/5} = \tfrac{-1-\sqrt{5}}{4}+\sqrt{\tfrac{5-\sqrt{5}}{8}}\,i$$
$$b = (e^{7\pi i/5})^{1/5} = e^{47\pi i/25} = e^{-3\pi i/25} = \left(\tfrac{1-\sqrt{5}}{4}-\sqrt{\tfrac{5+\sqrt{5}}{8}}\,i\right)^{1/5}$$
which should be prettier enough.
