Ever heard of complex numbers

For variety, we know that:

$\begin{aligned} \cos \left(\frac{\pi}{9}\right) + \color{#3D99F6}{\cos \left(\frac{3\pi}{9}\right)} + \cos \left(\frac{5\pi}{9}\right) + \cos \left(\frac{7\pi}{9}\right) & = \frac{1}{2} \quad \quad \small \color{#3D99F6}{\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}} \\ \cos \left(\frac{\pi}{9}\right) + \color{#3D99F6}{\frac{1}{2}} + \cos \left(\frac{5\pi}{9}\right) + \cos \left(\frac{7\pi}{9}\right) & = \frac{1}{2} \\ \cos \left(\frac{\pi}{9}\right) + \cos \left(\frac{5\pi}{9}\right) + \cos \left(\frac{7\pi}{9}\right) & = 0 \end{aligned}$

Let $a = \cos \left(\frac{\pi}{9}\right)$ , $b = \cos \left(\frac{5\pi}{9}\right)$ , and $c = \cos \left(\frac{7\pi}{9}\right)$ . Then $\color{#3D99F6}{a + b + c = 0}$ . From the identity $a^3+b^3+c^3 = (\color{#3D99F6}{a + b + c})(a^2+b^2+c^2) - (ab+bc+ca)(\color{#3D99F6}{a + b + c}) + 3abc$ , we have $a^3+b^3+c^3 = 3abc$ . Therefore,

$\cos \left(\frac{\pi}{9}\right) \cos \left(\frac{5\pi}{9}\right) \cos \left(\frac{7\pi}{9}\right) \\ = \frac{1}{3} \left(\cos^3 \left(\frac{\pi}{9}\right) + \cos^3 \left(\frac{5\pi}{9}\right) + \cos^3 \left(\frac{7\pi}{9}\right) \right) \\ = \frac{1}{3} \left(\frac{1}{4} \left[4\left(\cos^3 \left(\frac{\pi}{9}\right) + \cos^3 \left(\frac{5\pi}{9}\right) + \cos^3 \left(\frac{7\pi}{9}\right) \right) - 3(\color{#3D99F6}{0}) \right] \right) \\ = \frac{1}{3} \left(\frac{1}{4} \left[4\left(\cos^3 \left(\frac{\pi}{9}\right) + \cos^3 \left(\frac{5\pi}{9}\right) + \cos^3 \left(\frac{7\pi}{9}\right) \right) - 3\left(\color{#3D99F6}{\cos \left(\frac{\pi}{9}\right) + \cos \left(\frac{5\pi}{9}\right) + \cos \left(\frac{7\pi}{9}\right)}\right) \right] \right) \\ = \frac{1}{12} \left[\cos \left(\frac{\pi}{3}\right) + \cos \left(\frac{5\pi}{3}\right) + \cos \left(\frac{7\pi}{3}\right)\right] \quad \quad \small \color{#3D99F6}{\text{Note that }\cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta} \\ = \frac{1}{12} \left[ \frac{1}{2} + \frac{1}{2} + \frac{1}{2} \right] = \boxed{\frac{1}{8}}$

$\cos\ 20 \times \cos40 \times \cos80$ is the given expression . Now using $cos\theta \times \cos(60-\theta) \times \cos(60 +\theta) = 0.25 \times \cos(3\times \theta)$ .Putting $\theta$ = $20$ answer is $\boxed{0.125}$ . Note that all angles are in degrees

It's Morrie's Law!

A standard way to solve this problem is by using product to sum formulas.

$\cos(a) \cos(b) = \dfrac{1}{2} \left[\cos(a+b) + \cos(a-b) \right]$

$= \dfrac{1}{2} \left[ \cos\left(\dfrac{2\pi}{3}\right) + \cos \left(\dfrac{4\pi}{9}\right) \right] \cos \left(\dfrac{7\pi}{9} \right)$

$= \dfrac{1}{2}\cos\left( \dfrac{4\pi}{9}\right)\cos\left(\dfrac{7\pi}{9}\right) + \dfrac{1}{2}\cos\left(\dfrac{2\pi}{3}\right) \cos \left(\dfrac{7\pi}{9} \right)$

$= \dfrac{1}{4} \left[ \cos\left( \dfrac{11\pi}{9}\right) + \cos\left(\dfrac{\pi}{3}\right) \right] - \dfrac{1}{4} \cos \left(\dfrac{7\pi}{9} \right)$

Since cosine is an even function, we know that $\cos \left( \dfrac{11\pi}{9}\right) = \cos \left( \dfrac{7\pi}{9}\right)$ . We also know that $\cos \left( \dfrac{\pi}{3}\right) = \dfrac{1}{2}$ .

Simplifying, the answer is $\boxed{\dfrac{1}{8}}$