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. Demostración – Formulas de Cardano Vieta. lutfinn (48) in cardano • 5 months ago. source · cardano. 5 months ago by lutfinn (48). $ 1 vote. + lutfinn. N 1 N N. N) xi = \, i.e. of A TT (x-a;) = } II (x-ak) j=1 J j=1 – j=1 ifk From here we easily obtain, by the Cardano-Vieta relations, N N) N N N y: = + +) as. Hence.
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Results 1 to 4 of 4. Cardano-Vieta, cubics roots and i.
Vieta and Cardano
Cardano-Vieta, cubics roots and i Whats up! 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. And in order to solve this ecuation for cos aI used the first Cardano-Vieta s formula, as trigonometric ratios are real numbers, and in the two other ecuations there is an i.
Im sure I havent had a mistake, because the calculator says its right. Thank you all ccardano sorry for my bad grammar, Im Spanish. Last edited by gines; Jan 23rd at Cardano-Vieta, cubics roots and i So you do not know how to take the cube root of a complex number?
Similarly for the other cube root.
Vieta’s formulas – Wikipedia
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.
What did you get for that? Last edited by HallsofIvy; Jan 23rd at Apr 1st Apr 2nd Jan 27th Sep 27th Cardano’s Formula – please help Posted in the Algebra Forum. Oct 3rd ,