Angles In a Pentagon

$\begin{aligned} X & = \sec \frac \pi{10} \sec \frac {3\pi}{10} \sec \frac {7\pi}{10} \sec \frac {9\pi}{10} \\ \implies \frac 1X & = \cos \frac \pi{10} \cos \frac {3\pi}{10} \color{#3D99F6} \cos \frac {7\pi}{10} \cos \frac {9\pi}{10} & \small \color{#3D99F6} \text{Note that} \cos (\pi - x) = - \cos x \\ & = \cos \frac \pi{10} \cos \frac {3\pi}{10} \color{#3D99F6} \left( - \cos \frac {3\pi}{10}\right)\left(- \cos \frac \pi{10}\right) \\ & = \cos^2 \frac \pi{10} \cos^2 \frac {3\pi}{10} \\ & = \frac 14 \left(1+ \cos \frac \pi 5 \right) \left(1 + \cos \frac {3\pi}5\right) \\ & = \frac 14 \left(1+ \cos \frac \pi 5 + \cos \frac {3\pi}5 + {\color{#3D99F6}\cos \frac \pi 5 \cos \frac {3\pi} 5} \right) & \small \color{#3D99F6} \text{As } \cos A \cos B = \frac 12 (\cos (A-B) + \cos (A+B)) \\ & = \frac 14 \left(1+ \cos \frac \pi 5 + \cos \frac {3\pi}5 + {\color{#3D99F6} \frac 12 \left( \cos \frac {2\pi}5 + \cos \frac {6\pi} 5 \right)} \right) & \small \color{#3D99F6} \text{Note that} \cos (\pi - x) = - \cos x \\ & = \frac 14 \left(1+ \cos \frac \pi 5 + \cos \frac {3\pi}5 - \frac 12 \left({\color{#3D99F6} \cos \frac {3\pi}5 + \cos \frac \pi 5} \right) \right) \\ & = \frac 14 \left(1+ \frac 12 \left({\color{#3D99F6}\cos \frac \pi 5 + \cos \frac {3\pi} 5}\right) \right) & \small \color{#3D99F6} \text{See note: } \sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) = \frac 12 \\ & = \frac 14 \left(1+ \frac 12 \left({\color{#3D99F6}\frac 12}\right) \right) \\ & = \frac 5{16} \\ \implies X & = \frac {16}5 = \boxed{3.2} \end{aligned}$

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Note:

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

My solution using De Moivre's Theorem:

We will use the expansion of $\cos 5\theta$ to solve this problem.

$\begin{aligned} \text{cis} 5 \theta &= (\text{cis} \theta)^5\\ \cos 5\theta + i \sin 5\theta &= (\cos \theta + i \sin \theta)^5\\ &= (\cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta) + (\cos^4 \theta \sin \theta - \cos^2 \theta \sin^3 \theta) i\\ \end{aligned}$

Equating real parts and converting $\sin$ to $\cos$ gets us

$\begin{aligned} \cos 5\theta &= \cos^5 \theta - 10 \cos^3 \theta (1 - \cos^2 \theta) + 5 \cos \theta (1-cos^2 \theta)^2 \\ &= 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta \\ \end{aligned}$

Notice that $\frac {\pi}{10}, \frac {3\pi}{10}, \frac {5\pi}{10}, \frac {7\pi}{10}, \frac {9\pi}{10}$ are the 5 roots to $\cos 5 \theta = 0$ . (This is fairly simple to show and is left as an exercise to the reader). However, $\cos \frac {5\pi}{10} = 0$ , so we can divide out this root and get polynomial

$16 \cos^4 \theta - 20 \cos^2 \theta + 5 = 0$

By Vieta's, the product of the other roots is $\frac {5}{16}$ , so $\cos \frac {\pi}{10} \cos \frac {3\pi}{10} \cos \frac {7\pi}{10} \cos \frac {9\pi}{10} = \frac {5}{16}$ . From this, we get

$\sec \dfrac {\pi}{10} \sec \dfrac {3\pi}{10} \sec \dfrac {7\pi}{10} \sec \dfrac {9\pi}{10} = \dfrac {16}{5} = 3.2$