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I want to show that area of a circle $D$ with radius $r$ is $\pi r^2$. To do that I consider inscribed and circumscribed regular $n$-gons $I_n$ and $O_n$. Their areas equals to $$ S(I_n) = \frac 12 n r^2 \sin\frac{2\pi}n \quad\text{and}\quad S(O_n) = nr^2 \tan \frac\pi n. $$ As $I_n \subset D \subset O_n$, it follows that $$ \frac 12 r^2 n\sin\frac{2\pi}n \leq S(D) \leq r^2 n\tan \frac\pi n. $$ Then, by passage to limits, $$ \frac 12 r^2 \cdot 2\pi \leq S(D) \leq r^2 \cdot\pi. $$ However, I need the knowledge that $\lim_{n\to\infty} n\sin (1/n) = 1$. Luckily, I know that for $x \in (0, \pi/2)$ $$ \sin x < x < \tan x. $$ But the classic proof of a right inequality requires calculation of area of a circle segment! What to do?

I am not sure that proofs by integration do not use the derivative of $\sin x$ (which is, again, follows from a limit $\lim_{x\to0} (\sin x)/x$). Same thing about tayloring to prove inequalities.

Shoul I somehow show that arc length is less than length of a tangent segment?

UPD Now I see two ways, both using definition of length of a curve as supremum of inscribed polygonal chains.

One is to derive inequalty $x < \tan(x)$ from the fact that if $A \subset B$ are convex polygons, then the perimeter of $A$ is at most the perimiter of $B$. (Thanks to @Mark S.)

Second is to consider a sequence of polygons inscribed in circle whose perimeters converge to $2\pi r$. Area of each such polygon is $$ \frac12 \sum a_i h_i, $$ where $a_i$ are sides' lengths and $h_i$ are distances from sides to the center of circle. As perimeter of polygon tends to the perimeter of circle, $\max a_i$ tends to 0, so $\min h_i$ and $\max h_i$ tend to $r$.

G. Strukov
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    I'm not sure I understand exactly what you're asking, but you may be interested in Richman's brief write-up A Circular Argument. – Mark S. Nov 13 '20 at 02:10
  • On a circle with radius 1 and angle x draw sin(x) and tag(x). That x is larger than sin(x) is straight forward. With tag(x) you need somewoek. – Moti Nov 13 '20 at 05:10
  • @moti, yes, I know that, but how exactly I show that $x < \tan(x)$ without cirlce area and sin derivative? – G. Strukov Nov 13 '20 at 05:34
  • You need to first have a clear understanding of the meaning of symbols $\pi, \sin x, \cos x, \tan x$ given $x$ a real number. When you do this precisely then you will find that there is no circularity involved in finding area of circle. A typical definition of $\pi$ is the area of unit circle (or half the circumference of a unit circle). – Paramanand Singh Nov 13 '20 at 07:20
  • Please update your question with the relevant definitions of the symbols mentioned in my last comment. Typical high school texts ignore such fine details and assume that the student will somehow get the hang of it. Unfortunately by the time you get the details as part of some undergrad curriculum all the interest is lost. – Paramanand Singh Nov 13 '20 at 07:20
  • @ParamanandSingh I define $\pi$ as the perimeter of circle with diameter $1$. Trigonometric functions defined as ratios of sides of a right triangle. – G. Strukov Nov 13 '20 at 07:38
  • @G.Strukov: your definition of $\pi$ is fine. But the definition of trigonometric ratios does not really help in giving a link between $x$ and $\sin x$. The $x$ simply does not appear in the triangle. However with your definition of $\pi$ you can prove that area of a circle of unit radius is $\pi$. – Paramanand Singh Nov 13 '20 at 07:49
  • See https://math.stackexchange.com/a/1567245/72031 – Paramanand Singh Nov 13 '20 at 08:05
  • The perimeter of a circle with radius 1 is $2\pi$ – Moti Nov 14 '20 at 06:47

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Comment:

$Lim_{n\rightarrow\infty}sin (\alpha)=lim_{n\rightarrow \infty}\frac{\alpha}n=d\alpha$

So: $dA=\frac 12 \cdot r\cdot r\cdot d\alpha=\frac 12 \cdot r^2 d\alpha$

$A=\frac12 r^2\int^{2\pi}_0 \cdot d\alpha=\frac 12 r^2\cdot \big[\alpha\big]^{2\pi}_0=\pi r^2$

sirous
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