Let's salute cos 2

Let $P$ be the expression, then we have:

$\begin{aligned} P & = \left(1 + \cos \frac{\pi}{10} \right) \left(1 + \cos \frac{3\pi}{10} \right) \left(1 \color{#3D99F6}{+ \cos \frac{7\pi}{10}} \right) \left(1 \color{#3D99F6}{+ \cos \frac{9\pi}{10}} \right) \quad \quad \small \color{#3D99F6} {\left[\cos \frac{k\pi}{10} = -\cos \left(\frac{10-k\pi}{10}\right) \right]} \\ & = \left(1 + \cos \frac{\pi}{10} \right) \left(1 + \cos \frac{3\pi}{10} \right) \left(1 \color{#3D99F6}{- \cos \frac{3\pi}{10}} \right) \left(1 \color{#3D99F6}{- \cos \frac{\pi}{10}} \right) \\ & = \left(1 - \cos^2 \frac{\pi}{10} \right) \left(1 - \cos^2 \frac{3\pi}{10} \right) \\ & = \sin^2 \frac{\pi}{10} \sin^2 \frac{3\pi}{10} \\ & = \frac{1}{4} \left(1 - \cos \frac{\pi}{5} \right) \left(1 - \color{#3D99F6}{\cos \frac{3\pi}{5}} \right) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6} {\left[ \cos \frac{\pi}{5} + \cos \frac{3\pi}{5} = \frac{1}{2}\right]} \\ & = \frac{1}{4} \left(1 - \cos \frac{\pi}{5} \right) \left(1 - \color{#3D99F6}{\left[ \frac{1}{2} - \cos \frac{\pi}{5} \right]} \right) \\ & = \frac{1}{8} \left(1 - \cos \frac{\pi}{5} \right) \left(1 + 2 \cos \frac{\pi}{5} \right) \\ & = \frac{1}{8} \left(1 + \cos \frac{\pi}{5} - \color{#3D99F6}{2 \cos^2 \frac{\pi}{5}} \right) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6} {\left[ 2 \cos^2 \frac{\pi}{5} - 1 = \cos \frac{2\pi}{5} \right]} \\ & = \frac{1}{8} \left( \cos \frac{\pi}{5} \color{#3D99F6}{-\cos \frac{2\pi}{5}} \right) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6} {\left[ \cos \frac{k\pi}{5} = - \cos \left(\frac{5-k\pi}{5}\right) \right]} \\ & = \frac{1}{8} \left( \cos \frac{\pi}{5} \color{#3D99F6}{+ \cos \frac{3\pi}{5}} \right) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6} {\left[ \cos \frac{\pi}{5} + \cos \frac{3\pi}{5} = \frac{1}{2}\right]} \\ & = \frac{1}{8} \left( \frac{1}{2} \right) = \frac{1}{16} \\ & \\ \Rightarrow \frac{1}{P} & = \boxed{16} \end{aligned}$

Consider the equation $\cos(5x) = 0$ , it has roots of $5x = \frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2},\frac{9\pi}{2}$ . Equivalently, it has roots of $x = \frac{\pi}{10},\frac{3\pi}{10},\frac{5\pi}{10},\frac{7\pi}{10},\frac{9\pi}{10}$ .

By quintuple angle formula,

$0 = \cos(5x) = 16\cos^5(x) - 20\cos^3(x) + 5\cos(x)$

has roots $x = \frac{\pi}{10},\frac{3\pi}{10},\frac{5\pi}{10},\frac{7\pi}{10},\frac{9\pi}{10}$ .

For simplicity sake, let $y = \cos(x)$ , then the equation

$16y^5 - 20y^3 + 5y = 0$

has roots $y = \cos\left(\frac{\pi}{10} \right), \cos\left(\frac{3\pi}{10}\right), \cos\left(\frac{5\pi}{10}\right), \cos\left( \frac{7\pi}{10}\right),\cos\left(\frac{9\pi}{10}\right)$ .

Let $y = Y - 1$ , then the equation

$16(Y-1)^5 - 20(Y-1)^3 + 5(Y-1) = 0$

has roots $Y = 1+ \cos\left(\frac{\pi}{10} \right),1+ \cos\left(\frac{3\pi}{10}\right), 1+\cos\left(\frac{5\pi}{10}\right),1+ \cos\left( \frac{7\pi}{10}\right),1+\cos\left(\frac{9\pi}{10}\right)$ .

By Vieta's formula,

$\begin{aligned} && \left [ 1+ \cos\left(\frac{\pi}{10} \right)\right ] \times \left [ 1+ \cos\left(\frac{3\pi}{10}\right) \right] \times \left [ 1+\cos\left(\frac{5\pi}{10}\right)\right ] \times \left [ 1+ \cos\left( \frac{7\pi}{10}\right) \right ] \times \left [ 1+\cos\left(\frac{9\pi}{10}\right) \right ] \\ && = - \frac{-16 + 20 - 5}{16} \\ && = \frac1{16}. \end{aligned}$

With $\cos\left(\frac{5\pi}{10}\right) = 0$ , we obtained the desired value of $\dfrac1{16}$ .