MHB Challenge: Is cos(pi/60) transcendental?

[Prove It]

Here's your challenge - is $\displaystyle \begin{align*} \cos{ \left( \frac{\pi}{60} \right) } \end{align*}$ transcendental, or does it have an exact surd value? If it has an exact surd value, what is it?

Spoiler

Here is my solution for those of us playing at home.

It can be shown that $\displaystyle \begin{align*} \cos{ \left( \frac{2\,\pi}{5} \right) } = \frac{\sqrt{5} - 1}{4} \end{align*}$, $\displaystyle \begin{align*} \sin{ \left( \frac{2\,\pi}{5} \right) } = \frac{\sqrt{10 + 2\,\sqrt{5}}}{4} \end{align*}$, $\displaystyle \begin{align*} \cos{ \left( \frac{\pi}{3} \right) } = \frac{1}{2} \end{align*}$ and $\displaystyle \begin{align*} \sin{ \left( \frac{\pi}{3} \right) } = \frac{\sqrt{3}}{2} \end{align*}$, so that means

$\displaystyle \begin{align*} \cos{ \left( \frac{\pi}{15} \right) } &= \cos{ \left( \frac{2\,\pi}{5} - \frac{\pi}{3} \right) } \\ &= \cos{ \left( \frac{2\,\pi}{5} \right) } \cos{ \left( \frac{\pi}{3} \right) } + \sin{ \left( \frac{ 2\,\pi}{5} \right) } \sin{ \left( \frac{\pi}{3} \right) } \\ &= \frac{ \left( \sqrt{5} - 1 \right) }{4} \cdot \frac{1}{2} + \frac{\sqrt{10 + 2\,\sqrt{5}}}{4} \cdot \frac{\sqrt{3}}{2} \\ &= \frac{\sqrt{5} - 1 + \sqrt{30 + 6\,\sqrt{5}}}{8} \end{align*}$

Now we should note that for angles in the first quadrant

$\displaystyle \begin{align*} \cos{ \left( \frac{\theta}{2} \right) } \equiv \sqrt{ \frac{ \cos{ \left( \theta \right) } + 1 }{2} } \end{align*}$

so

$\displaystyle \begin{align*} \cos{ \left( \frac{\pi}{30} \right) } &= \sqrt{ \frac{\frac{\sqrt{5} - 1 + \sqrt{30 + 6\,\sqrt{5}}}{8} + 1}{2} } \\ &= \sqrt{ \frac{\sqrt{5} + 7 + \sqrt{30 + 6\,\sqrt{5}}}{16} } \\ &= \frac{\sqrt{\sqrt{5} + 7 + \sqrt{30 + 6\,\sqrt{5}}}}{4} \\ \\ \cos{ \left( \frac{\pi}{60} \right) } &= \sqrt{ \frac{\frac{\sqrt{ \sqrt{5} + 7 + \sqrt{30 + 6\,\sqrt{5}} }}{4} + 1}{2} } \\ &= \sqrt{ \frac{\frac{ \sqrt{\sqrt{5} + 7 + \sqrt{30 + 6\,\sqrt{5}}} + 4}{4}}{2} } \\ &= \sqrt{\frac{\sqrt{\sqrt{5} + 7 + \sqrt{30 + 6\,\sqrt{5}}} + 4}{8}} \\ &= \sqrt{ \frac{2\,\sqrt{ \sqrt{5} + 7 + \sqrt{30 + 6\,\sqrt{5}} } + 8}{16} } \\ &= \frac{\sqrt{2\,\sqrt{\sqrt{5} + 7 + \sqrt{30 + 6\,\sqrt{5}}} + 8}}{4} \end{align*}$

So there you go, it has an exact surd value, disgusting as it is. As for whether it can be simplified further, I am unsure :)

 

[I like Serena]

Spoiler

Looks like an actual solution will be the root of a polynomial of order 16.
That is, algebraic, but probably not a 'nice' surd value.