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In the Squeeze Theorem proof of $\lim\limits_{\theta\to 0}\frac{\sin{\theta}}{\theta} $, the appeal is made to the obvious triangle inclusion in the image on the unit circle below to say that

$$\frac{\sin{\theta}}{2} \leq \frac{\theta}{2} \leq \frac{\tan{\theta}}{2} $$

While it is indeed obvious that these triangles share this inequality relationship, I was wondering how it can be rigorously proven. For the purposes of answering, it's fine to prove it for just the upper right quadrant of the unit circle, i.e. when $0 < \theta < \frac{\pi}{2}$.

enter image description here

N. F. Taussig
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Ben G
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  • I mean, this is a rigorous proof – Andrea B. Dec 10 '21 at 03:56
  • @AmejiB. : No, it's not. $\qquad$ – Michael Hardy Dec 10 '21 at 03:57
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    You want a rigorous proof of Area($\triangle ABC$)$,<,$Area(blue sector)$,<,$Area($\triangle ABD$)? Is that what you're asking? That'd have to start with a rigorous definition of the concept of area. Things here can get pretty complicated. – bjorn93 Dec 10 '21 at 03:57
  • @MichaelHardy What about it is not rigorous? I mean as bjorn points out, questions about area can get fundamentally messy, but you might as well be reinventing the wheel at that point – Andrea B. Dec 10 '21 at 04:00
  • @AmejiB. : A logically rigorous proof is amenable to algorithmic validity checking. – Michael Hardy Dec 10 '21 at 04:02
  • You did not answer why this is not logically rigorous – Andrea B. Dec 10 '21 at 04:02
  • @AmejiB. : Logical rigor is concerned with definitions and connectives of propositional logic and such things as the equvialence of $\lnot\exists x\cdots$ and $\forall x,\lnot\cdots$ and so on. – Michael Hardy Dec 10 '21 at 04:05
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    See my answer here for a slightly different take, based on the two assumptions 1) circular sectors are convex, 2) the perimeter of a convex closed curve is smaller than that of any other one enclosing it. – dxiv Dec 10 '21 at 04:09
  • You have still not answered my question... – Andrea B. Dec 10 '21 at 04:15
  • Thanks @dvix. Do those two assumptions follow from particular axioms? – Ben G Dec 10 '21 at 04:43
  • For a rigorous treatment of trigonometric functions we have to use Calculus, not Geometry. – Kavi Rama Murthy Dec 10 '21 at 05:00
  • @bgcode Those are basic properties that hold true in formalisms that define convexity and curve length. You don't say in the question which axioms you work with, or what you are willing to accept as "rigorous". – dxiv Dec 10 '21 at 05:26
  • I see @dvix. How about working from Euclid's axioms, or Hilbert's? Or using set theory axioms from ZFC without geometry axioms at all, if that's possible. I'm a math n00b, so I'm not sure which axioms I should assume to be working from. For example, in my earlier geometry notes, I wrote that Corresponding Angles Axiom was: If a transversal intersects two parallel lines, then each pair of corresponding angles are congruent. Not sure if that's a Euclidean axiom or what.. But I'd like to know the axioms I could/should learn/work with for being rigorous as I learn calc. – Ben G Dec 10 '21 at 05:44
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    @bgcode There is a long way from Euclid's axioms to rigorous definitions of both trig functions and geometric imagery (curves, convexity, arc length). You can find some pointers in related questions 1, 2, 3. – dxiv Dec 10 '21 at 06:03
  • @dxiv So it seems like I need to either accept some higher level axioms, or wait until I understand a lot more math to build up from those axioms? This is one of the first cases I've run into where something far from Euclid needed to be used as a geometric axiom, but maybe that's par for the course? Also one idea that came to mind is if you had a function defining the area of the sector, could you maybe set up an inequality from that? Just an idea. Any tips on how to approach rigor in general as I learn are welcome as well. Thank you! – Ben G Dec 10 '21 at 06:14
  • This follows from the Taylor series expansion of sine, since we need only check $\theta < 1$, and once you get the left-hand inequality, the right follows immediately. – While I Am Dec 10 '21 at 06:15

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