\(ABCDE\) is a regular pentagon. By projecting the broken line \(AED\) on the line \(AB\), or otherwise, show that \[\cos\dfrac{\pi}{5}-\cos\dfrac{2\pi}{5}=\dfrac{1}{2}.\]
Let’s assume that the side lengths of the pentagon are \(1\).
Can we draw a large, clear diagram of the pentagon \(ABCDE\), with \(AB\) horizontally at the bottom?
What are the angles inside a regular pentagon, in radians?
If we now drop perpendiculars from \(D\) and \(E\) to the horizontal, can we start to find lengths and angles?
Hence, or otherwise, show that \(\cos\dfrac{\pi}{5} = \dfrac{\sqrt{5}+1}{4}\).
Using the first part, are there any trigonometrical identities we could use? To get this just in terms of \(\cos\dfrac{\pi}{5}\)?
Show further that \(\cos\dfrac{3\pi}{5} = -\dfrac{\sqrt{5}-1}{4}\).
What’s the sum of \(\dfrac{3\pi}{5}\) and \(\dfrac{2\pi}{5}\)? How does this connect \(\cos\dfrac{3\pi}{5}\) and \(\cos\dfrac{2\pi}{5}\)?
