Minimal Polynomial Thing 11

Here is a bash:

$\sin\left(\frac{\pi}{5}\right)+\cos\left(\frac{\pi}{5}\right)+\tan\left(\frac{\pi}{5}\right)=\frac{1+\sqrt{\sec^2\left(\frac{\pi}{5}\right)-1}}{\sec\left(\frac{\pi}{5}\right)}+\sqrt{\sec^2\left(\frac{\pi}{5}\right)-1}$

(squareroots may be plus or minus but this does not affect the final answer, so we assume plus)

$\left(1+\sec\left(\frac{\pi}{5}\right)\right)^2=1+\left(\sec^2\left(\frac{\pi}{5}\right)+2\sec\left(\frac{\pi}{5}\right)\right)=1+\frac{1+2\cos\left(\frac{\pi}{5}\right)}{\cos^2\left(\frac{\pi}{5}\right)}$

But $1+2\cos\left(\frac{\pi}{5}\right)=4\cos^2\left(\frac{\pi}{5}\right)$ since

$\sum _ {k=0}^4 e^{\frac{2\pi i k}{5}}=0$

which implies

$-1=\Re\left(e^{-\frac{1}{5} (2 i\pi)}+e^{\frac{2 i\pi}{5}}+e^{-\frac{1}{5} (4 i\pi)}+e^{\frac{4 i\pi}{5}}\right)=\cos\left(\frac{4\pi}{5}\right)+\cos\left(\frac{2\pi}{5}\right)+\cos\left(\frac{2\pi}{5}\right)+\cos\left(-\frac{1}{5} (4\pi)\right)=2\cos\left(\frac{4\pi}{5}\right)+2\cos\left(\frac{2\pi}{5}\right)=2\cos\left(\frac{2\pi}{5}\right)-2\cos\left(\frac{\pi}{5}\right)=2\left(\cos\left(\frac{2\pi}{5}\right)-\cos\left(\frac{\pi}{5}\right)\right)=-1+2\cos^2\left(\frac{\pi}{5}\right)-\cos\left(\frac{\pi}{5}\right)$

So $\left(1+\sec\left(\frac{\pi}{5}\right)\right)^2=1+4=5$ , and $\sec\left(\frac{\pi}{5}\right)=\sqrt{5}-1$ . Hence,

$z=\sin\left(\frac{\pi}{5}\right)+\cos\left(\frac{\pi}{5}\right)+\tan\left(\frac{\pi}{5}\right)=\sqrt{\left(\sqrt{5}-1\right)^2-1}+\frac{\sqrt{\left(\sqrt{5}-1\right)^2-1}+1}{\sqrt{5}-1}=\frac{1}{4}\left(\sqrt{5}+\sqrt{50-10\sqrt{5}}+1\right)$

Hence

$64 z^4-64 z^3-336 z^2+176 z+484=\left(-8 z^2+4 z+22\right)^2=\left(\sqrt{5} (6-4 z)\right)^2=80 z^2-240 z+180$

So

$0=64\left(z^4-z^3-\frac{13 z^2}{2}+\frac{13 z}{2}+\frac{19}{4}\right)=64 p (z)$

$p(1)$ gives the answer

It is easy to check that $p(x)$ is minimal. The rational root theorem easily shows that $p(x)$ has no rational roots. A simple argument also shows that $p(x)$ does not factor into two quadratics. So $p(x)$ is irreducible.