Quartic eqn solutions for Cos/Sin

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SUMMARY

The discussion centers on finding special case solutions to the quartic equation x^4 + a x^3 + b x^2 + c x + d = 0 when x is defined as Cos[theta]. Key insights include the application of the half-angle identity, which can reduce the degree of the equation but introduces cosines of different arguments. It is established that if theta is a rational multiple of pi, then x=cos(theta) is algebraic; otherwise, it is transcendental and not a solution of any polynomial with algebraic coefficients. The limitations of these properties in simplifying the quartic equation are emphasized.

PREREQUISITES
  • Understanding of quartic equations and their general forms
  • Familiarity with trigonometric identities, specifically the half-angle identity
  • Knowledge of algebraic and transcendental numbers
  • Basic principles of Vieta's formulas and their applications
NEXT STEPS
  • Research the implications of the half-angle identity in polynomial equations
  • Study the properties of algebraic vs. transcendental numbers in depth
  • Explore the use of Vieta's formulas in solving quartic equations
  • Investigate the relationship between trigonometric functions and polynomial roots
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Mathematicians, students studying algebra and trigonometry, and anyone interested in the properties of polynomial equations involving trigonometric functions.

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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