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Thread: Cardano-Vieta, cubics roots and i

  1. #1
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    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
    >
    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.
    Last edited by gines000; Jan 23rd 2017 at 10:18 AM.
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  2. #2
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    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.
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    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.
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  4. #4
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    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?
    Last edited by HallsofIvy; Jan 23rd 2017 at 03:52 PM.
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