Online you can find a wide range of websites on the golden section.[br]Some of them assign a special status to the pentagon because the golden section appears in it. Here's definitely some confusion between the number [math]\varphi[/math] (or [math]\Phi[/math]) and the division of a segment according to the golden ratio. No one does assign a special status to the square because on exceptional metaphysical or esthetical grounds because the diagonal of a square with side 1 equals [math]\sqrt{2}[/math]. [br]Likewise people don't ask question about an equality as sin 60°= sin [math]\frac{\pi}{3}[/math]= [math]\frac{\sqrt{3}}{2}[/math].[br]Whell, there's an analog equality for the sine of [math]\frac{\pi}{10}[/math].[br]So no, there's nothing special on a pentagon. Starting from this equality it's even logic that in calculations in a pentagon the number [math]\varphi[/math] (and also [math]\Phi[/math]) occurs.[br]You might as well skip the prove, just remembering the message that there's nothing special on a pentagon or a pentagram only because '[i]it reveals the golden section[/i]'.[br]on the next pages you can find some illustrations. If you are not a mathematician, you can leave the numbers and formulas for what they are. Just remember that it's logic itself that you can find [math]\varphi[/math] and [math]\Phi[/math] in geometrical applications with pentagons and decagons.

The equality [math]sin\frac{\pi}{10}=\frac{\varphi}{2}[/math] is just a mathematical equality, just as [math]sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}[/math], you learn in trigonometry. [br]This means that it's obvious that you'll meet [math]\varphi[/math] where angles of [math]\frac{\pi}{10}=18°[/math] (or mutiples) occcur. After all the length of the side of a regular polygon with n angles and radius r equals [math]2rsin\frac{\pi}{n}[/math].[br]With the command SurdText you can easily check related equalities: [br][math]SurdText\left(cos\left(\frac{\pi}{5}\right)\right)=\frac{1+\sqrt{5}}{4}=\frac{\Phi}{2}[/math][br][math]SurdText\left(cos\left(\frac{3\pi}{5}\right)\right)=\frac{1-\sqrt{5}}{4}=-\frac{\varphi}{2}[/math][br][math]SurdText\left(sin\left(\frac{3\pi}{10}\right)\right)=\frac{1+\sqrt{5}}{4}=\frac{\Phi}{2}[/math]