What is the Range of the Given Expression When ##\sin 2\beta≠0##?

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Homework Help Overview

The discussion revolves around determining the range of the expression ##\displaystyle \frac{x^2+2x\cos\alpha+1}{x^2+2x\cos\beta+1}## under the condition that ##\sin 2\beta \neq 0##. Participants are exploring the implications of this condition and how it affects the behavior of the expression as ##x## varies.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss rewriting the expression and consider its limits as ##x## approaches infinity. There are attempts to identify maximum and minimum points analytically, with some expressing uncertainty about the feasibility of this approach. Others suggest breaking the problem into cases based on the values of ##\alpha## and ##\beta##.

Discussion Status

Some participants have proposed methods to analyze the expression, including rearranging it to form a quadratic equation and applying conditions on the discriminant. There is acknowledgment of the complexity introduced by the lack of limits on the ranges of ##\alpha## and ##\beta##, and suggestions to restrict these ranges for simplification.

Contextual Notes

Participants note the restriction that limits or calculus cannot be used in the problem-solving process. There is also mention of the potential triviality of cases where ##\cos(\alpha) = \cos(\beta)##.

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Homework Statement


If ##\beta## is such that ##\sin 2\beta≠0##, then show that the expression ##\displaystyle \frac{x^2+2x\cos\alpha+1}{x^2+2x\cos\beta+1}## (x real) always lies between ##\cos^2\alpha/\cos^2\beta## and ##\sin^2\alpha/\sin^2\beta##.


Homework Equations





The Attempt at a Solution


I can rewrite the given expression to
\frac{(x+\cos\alpha)^2+\sin^2\alpha}{(x+\cos\beta)^2+\sin^2\beta}
What should I do next? :confused:
 
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I think this expression has a single maximum and a single minimum (unless cos(α)=cos(β), but that special case is trivial)) and it has always the same limit for x-> +- infinity. I don't know if it is possible to find those maximal/minimal points in an analytic way, but it looks like an interesting approach.
 
mfb said:
I think this expression has a single maximum and a single minimum (unless cos(α)=cos(β), but that special case is trivial)) and it has always the same limit for x-> +- infinity. I don't know if it is possible to find those maximal/minimal points in an analytic way, but it looks like an interesting approach.

I can't use limits or calculus for this problem. :(
 
Hmm... but what if you know the maximum and minimum (by some "clever guessing" aka analysis) and you can show that all other values are smaller/larger?
 
Because you have to show it lies between two values of which, in general, either could be the larger, it will probably be easiest if you break it into separate cases on that basis.
Another complication is that there's no limit given on the ranges of alpha and beta. I would try to show that it suffices to prove it for those being in the range 0 to pi/2, say.
Having done that, I would try subtracting the lower of the presumptive bounds and try to show the result is positive, etc.
 
Okay so I tried this again but I couldn't use your hints. :(

Let the given expression be y. Rearranging,
x^2(1-y)+2x(\cos 2\alpha-y\cos 2\beta)+(1-y)=0
The discriminant must be positive here. Using this condition, I was able to prove that the range (y) lies between ##\cos^2\alpha/\cos^2\beta## and ##\sin^2\alpha/\sin^2\beta##. Is this a good way to solve the problem?
 
Pranav-Arora said:
Okay so I tried this again but I couldn't use your hints. :(

Let the given expression be y. Rearranging,
x^2(1-y)+2x(\cos 2\alpha-y\cos 2\beta)+(1-y)=0
The discriminant must be positive here. Using this condition, I was able to prove that the range (y) lies between ##\cos^2\alpha/\cos^2\beta## and ##\sin^2\alpha/\sin^2\beta##. Is this a good way to solve the problem?
Sounds much better than anything I thought of.
 

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