To prove that \(\theta=\frac{\pi}{5}\) satisfies \(\frac{1}{2}+\cos{(2\theta)}=\cos{\theta}\):

\(2\sin{\theta}(\frac{1}{2}+\cos{(2\theta)})=2\sin{\theta} \cos{\theta}\)

\(\sin{\theta}+ 2\sin{\theta} \cos{(2\theta)}=2\sin{\theta} \cos{\theta}\)

\(\sin{(2-1)\theta}+2\sin{\theta} \cos{(2\theta)}=\sin{(2\theta)}\)

\(\sin{(2\theta)}\cos{\theta}-\sin{\theta}\cos{(2\theta)}+2\sin{\theta} \cos{(2\theta)}=\sin{(2\theta)}\)

\(\sin{(2\theta)}\cos{\theta}+\sin{\theta}\cos{(2\theta)}=\sin{(2\theta)}\)

\(\sin{(2+1)\theta}=\sin{(2\theta)}\)

\(\sin{(3\theta)}=\sin{(2\theta)}\)

\(\sin{(3\theta)}=\sin{(\pi-2\theta)}\)

\(3\theta=\pi-2\theta\)

\((3+2)\theta=\pi\)

\(5\theta=\pi\)

\(\theta=\frac{\pi}{5}\)

**QED**

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TopNewestMore generally \(\displaystyle \sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) = \frac 12\) (see proof). This means \(\cos \frac \pi 5 + \cos \frac {3 \pi} 5 = \cos \frac \pi 5 - \cos \frac {2 \pi} 5 = \frac 12\)

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