The cube root of 1

After a week of doing a little bit of practice for the AMC 8, my older son has returned to Art of Problem Solving’s Precalculus book. The chapter he’s on know is about trig identities.

Unrelated to his work in that book, the cube root of 1 came up tonight and he said “that’s just 1, right?” So, we chatted . . .

First we talked about the equation $x^3 - 1 = 0$:

For the second part of the talk, we discussed the numbers $\frac{1}{2} \pm \frac{\sqrt{3}}{2}$ and their relation to the equation $e^{i\theta} = \cos(\theta) + i\sin(\theta)$

Finally, I connected the discussion with the double angle (and then the triple angle) formulas that he was learning today. You can use the same idea in this video with $\cos(5\theta)$ to find that $\cos(75^{o}) = (\sqrt{5} - 1)/4$:

So, a lucky comment from my son led to a fun discussion about some ideas from trig that he happened to be studying today 🙂