Radical expression for $\tan 14.4^\circ$ Given that
$\displaystyle\cos 20^\circ=\frac{\sqrt[3]{\frac{1-i\sqrt{3}}{2}}+\sqrt[3]{\frac{1+i\sqrt{3}}{2}}}{2}$
and
$\displaystyle\sin 20^\circ=\frac{i\left(\sqrt[3]{\frac{1-i\sqrt{3}}{2}}-\sqrt[3]{\frac{1+i\sqrt{3}}{2}}\right)}{2}$
It turns out that there is a nice radical expression for $\tan 20^\circ$:
$\displaystyle\tan 20^\circ=\sqrt{11-\left(1-i\sqrt{3}\right)\sqrt[3]{148+4i\sqrt{3}}-\left(1+i\sqrt{3}\right)\sqrt[3]{148-4i\sqrt{3}}}$.
My question is, how did one get from the first two expressions to the third? It's pretty straightforward to find the minimal polynomial for $\cos 20^\circ$ from the triple-angle formulas, which is $8x^3-6x-1=0$. In other words, where did the three "new" numbers ($11$ and $37\pm i\sqrt{3}$) in the expression come from?
Now a similar situation arises for $\tan 14.4^\circ$:
$\displaystyle\cos 14.4^\circ=\frac{\sqrt[5]{\frac{-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}}{4}}+\sqrt[5]{\frac{-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}}{4}}}{2}$
and likewise for $\sin 14.4^\circ$.
My gut instinct is that $\tan 14.4^\circ$ can be written as
$\tan 14.4^\circ=\sqrt{a-\left(-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}\right)\sqrt[5]{b+ci}-\left(-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}\right)\sqrt[5]{b-ci}}$
for some algebraic $a, b,$ and $c.$
My question is, what are the values of a, b, and c? And I'm betting that my "gut instinct" on $\tan 14.4^\circ$ is wrong, so if anyone could help me derive the radical expression for $\tan 14.4^\circ$, that would be great.

Try using:$$\tan(x)=\pm\frac{\sqrt{1-\cos^2(x)}}{\cos(x)}$$ because you already know $\cos(x),x=14.4^\circ$.
$$\cos 14.4^\circ=\frac{\sqrt[5]{\frac{-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}}{4}}+\sqrt[5]{\frac{-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}}{4}}}{2}$$
Therefore
$$\tan(14.4)=\tan\frac{2\pi}{25}= \frac{\sqrt{1-\bigg(\frac{\sqrt[5]{\frac{-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}}{4}}+\sqrt[5]{\frac{-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}}{4}}}{2}}\bigg)^2}{\frac{\sqrt[5]{\frac{-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}}{4}}+\sqrt[5]{\frac{-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}}{4}}}{2}} =  \frac{\sqrt{1-\bigg(\frac{\sqrt[5]{\frac{-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}}{4}}+\sqrt[5]{\frac{-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}}{4}}}{2}}\bigg)^2}{\sqrt[5]{-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}}+\sqrt[5]{-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}}}  $$
Good luck simplifying this. Here is proof of the answer. Please correct me and give me feedback!
