MHB 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 $$tan\frac{\pi}{5}=\sqrt{5-2\sqrt{5}}$$. Where did it come from? Does $$tan\frac{\pi}{n}$$ always equal $$\sqrt{n-2\sqrt{n}}$$? 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:

$x\equiv18^{\circ}$

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

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

 

[MarkFL]

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

$x\equiv18^{\circ}$

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

See if you can make progress from there...

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

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

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

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

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

Using a Pythagorean identity, we have:

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

Hence:

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

And so:

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

 

[Monoxdifly]

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

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

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

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.