Is there a non-geometic proof of this inequality

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    Inequality Proof
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Discussion Overview

The discussion centers around the inequality sin(x) ≤ x ≤ tan(x) and whether there exists a non-geometric proof of this inequality. Participants explore different approaches to proving the limit of sin(x)/x as x approaches 0, including analytical methods and series definitions of trigonometric functions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the necessity of the inequality sin(x) ≤ x ≤ tan(x) for proving the limit of sin(x)/x as x approaches 0, suggesting that the limit can be proven using the series definition of the sine function.
  • Another participant proposes defining sine and cosine through their McLaurin series or as solutions to specific differential equations, indicating that these definitions may eliminate the need for the inequality or limit in deriving the derivatives of these functions.
  • A further contribution mentions integrating cos(x) < 1 and 1 < (1/cos²(x)) to derive the inequalities, while noting that certain assumptions about the derivatives of trigonometric functions and properties of integration are necessary.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and methods of proving the inequality and the limit, indicating that multiple competing approaches exist without a clear consensus on the best method.

Contextual Notes

Some participants assume knowledge of derivatives of trigonometric functions and properties of integration, which may limit the applicability of their arguments to those familiar with these concepts.

Mu naught
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sinx ≤ x ≤ tanx

You need the above inequality to prove that lim x-> 0 of sinx/x = 1, but I've only ever seen it derived geometrically. Is there an analytical proof of the above inequality, from which you can continue the sinx/x proof as normal?
 
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that limit is elegantly proved using the series definition of the sin function. no need for that inequality.
 
As tauon suggests, it depends upon how you define the trig functions. You can define sine and cosine in terms of their McLaurin series (lots of fun proving periodicity from that!). You can also define "y= sin(x)" as the solution to the differential equation y"= -y with y(0)= 0, y'(0)= 1, and define "y= cos(x)" as the solution to the differential equation y"= -y with y(0)= 1, y'(0)= 0.

In either case, you don't need that inequality nor that limit to derive the derivatives of sine and cosine.
 
(lots of fun proving periodicity from that!)
All the horrible, horrible memories. None of it even came up in the Analysis paper. If you just want the inequalities, without being too fussy about "how much we know" at this point:
Integrate cosx<1 for the first bit.
Integrate 1<(1/cos^2 x) for the second bit.

(We have to assume we know the derivatives of trig functions, and a few properties of integration. From the proofs involved, we probably know enough to do the sinx/x anyway, but...there it is if you need to check the result in an exam or something).
 

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