Compute $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$ using roots of unity .

Question

Compute $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$.

The hint says to take $\zeta=\text {cos}(20^\circ)+ \text {i sin} (20^\circ)$ and then asks us to write $\cos(20^\circ)$ is terms of $\zeta$ .

Now, I took $\zeta=\text {cos}(20^\circ)+ \text {i sin} (20^\circ)$ be the 18 root of unit. So $\zeta^2=\text {cos}(40^\circ)+ \text {i sin} (40^\circ)$ and $\zeta^4=\text {cos}(80^\circ)+ \text {i sin} (80^\circ)$.

But I didn't understand , how to write $\cos(20^\circ)$ is terms of $\zeta$ .

Any hints ? Thanks in advance !

Answer 1

Notice that $\text{cis}(-20^{\circ}) = \cos(20^{\circ})-i\sin(20^{\circ})$, which is $e^{-20^{\circ}} = \frac{1}{\zeta}$. Now we can use elimination to get that $2\cos(20^{\circ}) = \zeta+\frac{1}{\zeta}$, and the rest follows from here. In particular, you can try multiplying both sides by $\zeta-\frac{1}{\zeta}$, which is identical to the standard trigonometric solution, but simple expansion works just as well using the fact that the 9th roots of unity sum to 0.

Answer 2

Try to substitute the following:

$$ \begin{align} 2\cos{(20)}&=\zeta-\zeta^{8}\\ 2\cos{(40)}&=\zeta^{2}-\zeta^{7}\\ 2\cos{(80)}&=\zeta^{4}-\zeta^{5}\\ \\ \zeta^{9}&=-1 \end{align} $$

Then use the equation

$$ \begin{align} \zeta^{8}-\zeta^{7}+\zeta^{6}-\zeta^{5}+\zeta^{4}-\zeta^{3}+\zeta^{2}-\zeta+1&=0 \end{align} $$

Answer 3

Here's an alternative method, which is rather neat as it only uses the double angle formulae which are quite basic. Here we go :) $$\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}$$ Applying this twice more, we have $$\sin x=2^3\sin\frac{x}{8}\cos\frac{x}{8}\cos\frac{x}{4}\cos\frac{x}{2}$$ You may see where this is going. Let $\frac{x}{2}=80$ degrees: $$\frac{x}{2}=80\implies x=160$$ $$\implies \sin160=2^3\sin20\cos20\cos40\cos80$$ But we know that $\sin20=\sin(180-20)=\sin160$ so we have $$\sin160=2^3\sin160\cos20\cos40\cos80$$ $$\implies2^3\cos20\cos40\cos80=1\implies \cos20\cos40\cos80=\frac{1}{8}$$ I hope you find that helpful or interesting, or maybe even both :))