Quartic eqn solutions for Cos/Sin

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The discussion focuses on finding special case solutions to the quartic equation x^4 + a x^3 + b x^2 + c x + d = 0 when x is expressed as Cos[theta]. Participants note that while Vieta's formulas provide properties of the roots, they may not simplify the equation effectively in this context. The half-angle identity is mentioned as a potential method to reduce the degree of the equation, but it introduces complexities with different cosine arguments. It is emphasized that if theta is a rational multiple of pi, x=cos(theta) is algebraic, while for irrational multiples, it becomes transcendental. Overall, the discussion highlights the challenges and limitations of simplifying quartic equations with cosine solutions.
natski
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Hi all,

I am looking for a special case solution to the quartic equation x^4 + a x^3 + b x^2 + c x + d = 0 in the case where x = Cos[theta]. Are there are any special properties of the solutions? For example, I know there are numerous properties from Vieta's formula but none of these really help simplify things if x=cosine.

Natski
 
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Depends on what you want to do, I suppose. The half-angle identity
\cos^2a=\frac{1+\cos(2a)}{2}
could reduce the degree, but at the cost of introducing cosines of different arguments. This may have been what you were referring to as not useful, though; I'm not sure.
 
I don't see how it can be considered much of a special case if x=cos\theta. All you're saying here is that -1\leq x\leq 1 which really doesn't help a great deal.
 
x=cos(\theta) is algebraic if \theta is a rational multiple of pi. I.e. it is the zero of a polynomial of integer coefficients.

It can be shown that if \theta is not a rational multiple of pi, then x=cos(\theta) is transcendental and thus not the solution of any polynomial of algebraic coefficients.
 
Mentallic said:
I don't see how it can be considered much of a special case if x=cos\theta. All you're saying here is that -1\leq x\leq 1 which really doesn't help a great deal.

I don't know... if the equation was 2ax^2-1=2a\cos^2(\theta)-1, I would consider \cos(2\theta) a simplification.
 
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