How Terrible Angles Become Nice With The Correct Combo

There is a general identity (see proof below):

$\sum_{k=1}^n \cos \left(\frac {2k-1}{2n+1}\pi \right) = \frac 12$

So when $n=2$ , we have:

$\begin{aligned} \cos \frac \pi 5 \blue{+ \cos \frac {3\pi}5} & = \frac 12 \\ \cos \frac \pi 5 \blue {-\cos \left(\pi - \frac {3\pi}5 \right)} & = \frac 12 \\ \cos \frac \pi 5 - \cos \frac {2\pi}5 & = \frac 12 \\ \implies \cos 36^\circ - \cos 72^\circ & = \boxed{0.5} \end{aligned}$

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Proof:

$\begin{aligned} S & = \sum_{k=1}^n \cos \left(\frac {2k-1}{2n+1}\pi \right) = \Re \left(\sum_{k=1}^n e^{\frac {2k-1}{2n+1}\pi i} \right) = \Re \left(e^{\frac {\pi i}{2n+1}}\sum_{k=0}^{n-1} e^{\frac {2k \pi i}{2n+1}} \right) \\ & = \Re \left(\frac {e^{\frac {\pi i}{2n+1}}\left(e^{\frac {2n\pi i}{2n+1}}-1\right)}{e^{\frac {2\pi i}{2n+1}}-1}\right) = \Re \left(\frac {e^{\pi i} - e^{\frac {\pi i}{2n+1}}}{e^{\frac {2\pi i}{2n+1}}-1}\right) = \Re \left(\frac {1+ e^{\frac {\pi i}{2n+1}}}{1 - e^{\frac {2\pi i}{2n+1}}}\right) \\ & = \Re \left(\frac 1{1-e^{\frac {\pi i}{2n+1}}} \right) = \Re \left(\frac 1{1-\cos \frac \pi{2n+1} - i\sin \frac \pi{2n+1}} \right) \\ & = \Re \left(\frac {1-\cos \frac \pi{2n+1} + i\sin \frac \pi{2n+1}}{\left(1-\cos \frac \pi{2n+1} - i\sin \frac \pi{2n+1}\right)\left(1-\cos \frac \pi{2n+1} + i\sin \frac \pi{2n+1}\right)} \right) \\ & = \Re \left(\frac {1-\cos \frac \pi{2n+1} + i\sin \frac \pi{2n+1}}{2-2\cos \frac \pi{2n+1}} \right) = \boxed{\frac 12} \end{aligned}$

Note that $\cos 36-\cos 72=\frac{2\cos^2 36-2\cos^2 72} {2\cos 36+2\cos 72}=\frac{\cos 72+1-\cos 144-1}{2\cos 36+2\cos 72}=\frac{1}{2}=0.5.$ This method is quite slick in my opinion! Source: 103 Trigonometric problems by Titu Andreescu.