Why is $5\tan(54^\circ) = \sqrt{25 + 10\sqrt{5}}$ and $\tan\left(\frac{\pi}{5}\right) = \sqrt{5 - 2\sqrt{5}}$?

Question

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!

Answer 1

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 $$ \tan\frac{3\pi}{10}=\cot\frac{\pi}{5}=\frac{1}{\sqrt{5-2\sqrt{5}}}= \sqrt{\frac{5+2\sqrt{5}}{25-20}}=\frac{\sqrt{25+10\sqrt{5}}}{5} $$

Answer 2

Thanks to @N.F.Taussig for pointing out some errors.

Suppose you have a regular pentagon, with vertices $\{A,B,C,D,E\}$ (labelled cyclically so $A$ is neighbor to $B$ and $E$). Let us take each side length to be $1$. Let $X$ be the length of a diagonal, say $AC$. Let's first compute $X$.

To do it, let $P$ be the intersection of $AC$ and $BE$. Now we do some easy angle chasing: $\angle{BCA}=36=\angle{BAC}$, $\angle{CBP}=72=\angle{CPB}$. $\angle{ABP}=36$. Of course $\Delta BPC$ is isosceles (though not similar) and this implies that $PC$ has length $1$ We also see that $\Delta ABP$ is isosceles with angles $108-36-36$. Thus it is similar to $\Delta ACB$. Similarity then lets us compute $X$: $$\frac {X-1}1=\frac 1X\implies X^2-X-1=0\implies X=\frac 12(1+\sqrt5)$$

Now drop the perpendicular from $P$ to $AB$, let $Q$ denote its foot. Clearly $\Delta PAQ$ is a right triangle with angles $90-36-54$. We know one leg ($\frac 12$) and the hypotenuse ($\frac 12 (\sqrt 5 -1 )$). That suffices to solve the triangle and gives you the answer you seek.

Answer 3

As $54\cdot5\equiv90\pmod{180},$

if $\tan5x=\infty,5x=180^\circ n+90^\circ$ where $n$ is any integer

$x=36^\circ n+18^\circ$ where $n\equiv0,\pm1,\pm2\pmod5$

Using $\tan5x$ expansion, the roots of $$5t^4-10t^2+1=0$$ are $\tan(36^\circ n+18^\circ)$ where $n\equiv0,\pm1,2\pmod5$

as $n=2\implies\tan(36^\circ n+18^\circ)=?$

$$t^2=\dfrac{10\pm4\sqrt5}{2\cdot5}=\dfrac{5\pm2\sqrt5}5$$

As $\tan54^\circ>\tan18^\circ>0,\tan^254^\circ>\tan^218^\circ,$

$\tan^254^\circ=\dfrac{5+2\sqrt5}5=\dfrac{25+10\sqrt5}{5^2}$ and $\tan^218^\circ=\dfrac{25-10\sqrt5}{5^2}$

For $\tan36^\circ,$ can you use the same idea?