Thread: Cardano-Vieta, cubics roots and i

1. Cardano-Vieta, cubics roots and i

Whats up! Im new here. I was trying to demonstrate that the trigonometric ratios of every single integer grade can be written in function of roots. There is a table of trigonometric ratios of multiples of 3 in function of roots. So, if I get to figure out the exact trigonometric ratios of 1 grade, i will have got the trigonometric ratios of all the integers, by addition. I used the ecuation cos (3a) = 4cos3 (a) - 3cos (a) . All I have to do, is call a=1 and do the operations. But, in order to simplify the operations, I supposed a=20, since cos (60) = 0.5. And in order to solve this ecuation for cos (a), I used the first Cardano-Vieta s formula, as trigonometric ratios are real numbers, and in the two other ecuations there is an i.
And finally, I get this ecuation
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How can I remove i? Im sure I havent had a mistake, because the calculator says its right.
Thank you all and sorry for my bad grammar, Im Spanish.

2. Re: Cardano-Vieta, cubics roots and i

So you do not know how to take the cube root of a complex number? Given a number a+ bi, write it in "polar form" $rcos(\theta)+ irsin(\theta)$ where $r= \sqrt{a^2+ b^2}$ and $\theta= arctan(b/a)$ (adding $2\pi$ to $\theta$ does not change the root). Then the cube roots are given by $r^{1/3}(cos()\theta/3)+ sin(\theta/3))$. In the first, $(1+ i\sqrt{3})^{1/3}$, a=1 and $b= \sqrt{3}$ so that $r= sqrt{1+ 3}= 2$ and $\theta= arctan(\sqrt{3}/1)= 30$ degrees. So the three cube roots of $1+ i\sqrt{3}$ are $2^{1/3}(cos(10)+ i sin(10))$, $2^{1/3}(cos(130)+ i sin(130))$ and $2^{1/3}(cos(250)+ i sin(250))$. Similarly for the other cube root.

3. Re: Cardano-Vieta, cubics roots and i

Thank you mate. I have already tried it, but as you can see, it isnt right. Maybe its because there are three diferents cubic roots for only a complex number, so you cant just go for this way, because there would be many solutions for this ecuation, what doesnt totally make sense. I have heard something related to expanding the binomial, but I actually dont know how to do it.

4. Re: Cardano-Vieta, cubics roots and i

You have $\sqrt[3]{2}(cos(10)+ isin(10)$, as I suggested, but you do not have the equivalent for $\sqrt[3]{1- i\sqrt{3}}$. What did you get for that?