From,
Z^6 +1= (z^2 -2zcospi/6 +1)(z^2 -2zcospi/2 +1)(z^2 -2zcos5pi/6 +1)
Deduce:
Cos3theta= 4(Costheta-cospi/6)(Costheta-cospi/2)(Costheta-cos5pi/6)
Hi, I wasn’t too sure how to do these sort of questions where it says from this deduce this. Like what are my boundaries. For example, am I allowed to work with the second equation (the one we have to deduce) or not?
Also apparently this is a De Moivres theorem question but I can’t seem to figure out why. If you could please give me a hint on where to start that would be great.
Thank you!!!!!
You should only assume stuff from the first equation. You can work with one side at a time for what you wish to prove, but that probably was not the intended approach of the question since it says "deduce".
For this question, as a hint the first step is to identify that:
\begin{align*}\frac{z^6+1}{z^3} &= \frac{z^2-2z\cos \frac\pi6+1}{z} \times \frac{z^2-2z\cos\frac\pi2 + 1}{z} \times \frac{z^2-2z\cos\frac{5\pi}6 + 1}{z} \\ \therefore z^3 + z^{-3} &= \left(z - 2\cos \frac\pi6 + z^{-1} \right)\left(z - 2 \cos \frac\pi2 + z^{-1} \right)\left( z - 2\cos \frac{5\pi}6 + z^{-1}\right)\end{align*}
Now think about what you can do to throw away all the \(z\)'s and \(z^{-1}\)'s for a bunch of cosines. And yes, it is a de Moivre's question