Is tan(36°) always equal to √(5-2√5) for regular pentagons?

[Monoxdifly]

I was reading about the area of regular pentagon and it said that \(\displaystyle tan\frac{\pi}{5}=\sqrt{5-2\sqrt{5}}\). Where did it come from? Does \(\displaystyle tan\frac{\pi}{n}\) always equal \(\displaystyle \sqrt{n-2\sqrt{n}}\)? If yes, what is the proof?

 

[MarkFL]

If I were trying to derive the value of $\tan\left(36^{\circ}\right)$, I would likely begin with:

$x\equiv18^{\circ}$

\(\displaystyle \sin(2x)=\cos(3x)\)

See if you can make progress from there...

 

[Monoxdifly]

I have got some hints from My Math Forum yesterday:

There's a proof that $\cos(36^\circ)=\dfrac{1+\sqrt5}{4}$ here. Use the identities
$$\sin(x)=\sqrt{1-\cos^2(x)}$$
($x$ is in the first quadrant) and
$$\tan(x)=\dfrac{\sin(x)}{\cos(x)}$$
to finish up, if you will.

Then I finished it up:
$$\tan(x)=\dfrac{\sin(x)}{\cos(x)}$$
\(\displaystyle =\frac{\sqrt{1-cos^2x}}{\frac{1+\sqrt{5}}{4}}\)
\(\displaystyle =\frac{\sqrt{1-(\frac{1+\sqrt{5}}{4})^2}}{\frac{1+\sqrt{5}}{4}}\)
\(\displaystyle =\frac{\sqrt{\frac{16}{16}-\frac{1+2\sqrt{5}+5}{16}}}{\frac{1+\sqrt{5}}{4}}\)
\(\displaystyle =\frac{\sqrt{\frac{16-1-2\sqrt{5}-5}{16}}}{\frac{1+\sqrt{5}}{4}}\)
\(\displaystyle =\frac{\sqrt{\frac{10-2\sqrt{5}}{16}}}{\frac{1+\sqrt{5}}{4}}\)
\(\displaystyle =\frac{\frac{\sqrt{10-2\sqrt{5}}}{4}}{\frac{1+\sqrt{5}}{4}}\)
\(\displaystyle =\frac{\sqrt{10-2\sqrt{5}}}{1+\sqrt{5}}\)
\(\displaystyle =\sqrt{\frac{{10-2\sqrt{5}}}{(1+\sqrt{5})^2}}\)
\(\displaystyle =\sqrt{\frac{10-2\sqrt{5}}{1+2\sqrt{5}+5}}\)
\(\displaystyle =\sqrt{\frac{{10-2\sqrt{5}}}{6+2\sqrt{5}}}\)
\(\displaystyle =\sqrt{\frac{{10-2\sqrt{5}}}{6+2\sqrt{5}}\cdot\frac{6-2\sqrt{5}}{6-2\sqrt{5}}}\)
\(\displaystyle =\sqrt{\frac{{60-12\sqrt{5}-20\sqrt{5}+4(5)}}{36-4(5)}}\)
\(\displaystyle =\sqrt{\frac{{60-32\sqrt{5}+20}}{36-20}}\)
\(\displaystyle =\sqrt{\frac{{40-32\sqrt{5}}}{16}}\)
\(\displaystyle =\sqrt{5-2\sqrt{5}}\)

 

[MarkFL]

If I were trying to derive the value of $\tan\left(36^{\circ}\right)$, I would likely begin with:

$x\equiv18^{\circ}$

\(\displaystyle \sin(2x)=\cos(3x)\)

See if you can make progress from there...

\(\displaystyle \sin(2x)=\cos(2x)\cos(x)-\sin(2x)\sin(x)\tag{1}\)

Using (1) we can get a quadratic in $\sin(x)$:

\(\displaystyle 2\sin(x)=1-4\sin^2(x)\)

\(\displaystyle 4\sin^2(x)+2\sin(x)-1=0\)

\(\displaystyle \sin(x)=\frac{-1+\sqrt{5}}{4}\)

Using a Pythagorean identity, we have:

\(\displaystyle \cos(x)=\sqrt{1-\sin^2(x)}=\frac{\sqrt{2(5+\sqrt{5})}}{4}\)

Hence:

\(\displaystyle \tan(x)=\sqrt{\frac{(\sqrt{5}-1)^2}{2(5+\sqrt{5})}}\)

And so:

\(\displaystyle \tan(2x)=\frac{2\sqrt{\dfrac{(\sqrt{5}-1)^2}{2(5+\sqrt{5})}}}{1-\dfrac{(\sqrt{5}-1)^2}{2(5+\sqrt{5})}}=\sqrt{\frac{\sqrt{5}(\sqrt{5}-1)^3}{8}}=\sqrt{\frac{8(5-2\sqrt{5})}{8}}=\sqrt{5-2\sqrt{5}}\)

 

[Monoxdifly]

\(\displaystyle \sin(2x)=\cos(2x)\cos(x)-\sin(2x)\sin(x)\tag{1}\)

Using (1) we can get a quadratic in $\sin(x)$:

\(\displaystyle 2\sin(x)=1-4\sin^2(x)\)

Which (1) are you talking about?

 

[Greg]

$$\sin(2x)=\cos(2x)\cos(x)-\sin(2x)\sin(x)$$
$$2\sin(x)\cos(x)=(1-2\sin^2(x))\cos(x)-2\sin^2(x)\cos(x)$$
$$2\sin(x)=1-4\sin^2(x)$$
$$4\sin^2(x)+2\sin(x)-1=0$$

 

[MarkFL]

Which (1) are you talking about?

There is only 1 (1) that I posted:

View attachment 5444

 

[Monoxdifly]

Oops, sorry. I didn't see it. It's so far on the right that it somehow escaped my peripheral vision.