<s>[INST] Why is $5\tan(54^\circ) = \sqrt{25 + 10\sqrt{5}}$ and $\tan\left(\frac{\pi}{5}\right) = \sqrt{5 - 2\sqrt{5}}$? On the Wikipedia Page about Pentagons, I noticed a statement in their work saying that $\sqrt{25+10\sqrt{5}}=5\tan(54^{\circ})$ and $\sqrt{5-2\sqrt{5}}=\tan(\frac {\pi}{5})$
My question is: How would you justify that? My goal is to simplify $\sqrt{25+10\sqrt{5}}$ and $\sqrt{5-2\sqrt{5}}$ without knowledge on the detested radical!
 [/INST] You surely know that $z=\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}$ is the root in the first quadrant of $z^5-1=0$ so it satisfies
$$
z^4+z^3+z^2+z+1=0
$$
that can also be written as
$$
z^2+\frac{1}{z^2}+z+\frac{1}{z}+1=0
$$
or, by noting that $z^2+1/z^2=(z+1/z)^2-2$,
$$
\left(z+\frac{1}{z}\right)^2+\left(z+\frac{1}{z}\right)-1=0
$$
Therefore,
$$
z+\frac{1}{z}=2\cos\frac{2\pi}{5}=\frac{\sqrt{5}-1}{2}
$$
Thus
$$
\cos\frac{2\pi}{5}=\frac{\sqrt{5}-1}{4},
\qquad
\sin\frac{2\pi}{5}=\sqrt{1-\frac{5-2\sqrt{5}+1}{16}}=
\frac{\sqrt{10+2\sqrt{5}}}{4}
$$
Now use that
$$
\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}
$$
and you get
$$
\tan\frac{\pi}{5}=\frac{\sqrt{10+2\sqrt{5}}}{3+\sqrt{5}}=
\frac{\sqrt{(10+2\sqrt{5})(3-\sqrt{5})^2}}{4}=
\frac{\sqrt{80-32\sqrt{5}}}{4}=\sqrt{5-2\sqrt{5}}
$$
Since $54^\circ=3\pi/10=\pi/2-\pi/5$, we have
$$
\tan\frac{3\pi}{10}=\cot\frac{\pi}{5}=\frac{1}{\sqrt{5-2\sqrt{5}}}=
\sqrt{\frac{5+2\sqrt{5}}{25-20}}=\frac{\sqrt{25+10\sqrt{5}}}{5}
$$
</s>