How to Estimate Pi Without Trig Functions

[kalikusu]

Have neither seen an estimation nor derivation of pi that does not use trig functions. This is problematic as trig functions require radian inputs, via the relation pi radians = 180 deg. But if looking for pi, then how to get the input for the trig functions without pi?

Sure there are calculators that allow degree inputs provided, the calculator is set for DEG mode. Internally, the degrees are converted to radians before the trig function is evaluated.

Using Geometry alone, namely equation for a circle, the tangent to any point on the circle intersecting with a normal line to determine the length of an inscribed or circumscribed polygon can be used but with problems.

If focus on polygons starting with 4 sides and doubling with each iteration starting with the normal y = x, then the angle between a chosen segment of the polygon will halve with each iteration. On a spreadsheet, the estimate for pi then approaches 4.

If start with the normal y = x to identify the mid-point of the chord for the inscribed polygon and the point on the tangent to the circumscribed polygon, then the next normal has to be determined via the tangent to that point. But if do this, then after the 15th iteration , the estimate for pi becomes erratic.

Tables

radius​

1​

For the inscribed polygon:​

For circumscribed polygon:​

diameter​

2​

Yt intersects with Yn​

Normal line intersects with circle​

Inscribed polygon x-intercept:​

x0​

1​

y0​

0​

b = y - x*Mt​

Inscribed​

Circumscribed​

Polygon​

with​

angle (deg)​

slope​

Tangent line​

x​

y​

u​

z​

circumference​

Pi - estimate​

x​

y​

u​

z​

circumference​

Pi - estimate​

order​

even​

360/sides​

tangent​

normal​

y-intercept​

b/(Mn-Mt)​

x*Mn​

x0 - x​

2*sqrt((y*y + u*u))​

sides * z​

circumference/diameter​

r/(sqrt(1+Mn*Mn))​

x*Mn​

x0 - x​

2*sqrt((y*y + u*u))​

sides * z​

circumference/diameter​

sides​

Mt​

Mn​

b = y0 - mx0​

2​

4​

90​

−1.0000000000​

1.00000000000000000000​

1.0000000000​

0.50000000​

0.50000000​

0.50000000​

1.41421356​

5.65685425​

2.82842712​

0.70710678​

0.70710678​

0.29289322​

1.53073373​

6.12293492​

3.0614674589​

3​

8​

45​

−2.4142135624​

0.41421356237309514547​

2.4142135624​

0.85355339​

0.35355339​

0.14644661​

0.76536686​

6.12293492​

3.06146746​

0.92387953​

0.38268343​

0.07612047​

0.78036129​

6.24289030​

3.1214451523​

4​

16​

22.5​

−5.0273394921​

0.19891236737965800607​

5.0273394921​

0.96193977​

0.19134172​

0.03806023​

0.39018064​

6.24289030​

3.12144515​

0.98078528​

0.19509032​

0.01921472​

0.39206856​

6.27309698​

3.1365484905​

5​

32​

11.25​

−10.1531703876​

0.09849140335716434491​

10.1531703876​

0.99039264​

0.09754516​

0.00960736​

0.19603428​

6.27309698​

3.13654849​

0.99518473​

0.09801714​

0.00481527​

0.19627070​

6.28066231​

3.1403311570​

6​

64​

5.625​

−20.3554676250​

0.04912684976946677523​

20.3554676250​

0.99759236​

0.04900857​

0.00240764​

0.09813535​

6.28066231​

3.14033116​

0.99879546​

0.04906767​

0.00120454​

0.09816491​

6.28255450​

3.1412772509​

7​

128​

2.8125​

−40.7354838721​

0.02454862210892543375​

40.7354838721​

0.99939773​

0.02453384​

0.00060227​

0.04908246​

6.28255450​

3.14127725​

0.99969882​

0.02454123​

0.00030118​

0.04908615​

6.28302760​

3.1415138011​

8​

256​

1.40625​

−81.4832402065​

0.01227246237956959064​

81.4832402065​

0.99984941​

0.01227061​

0.00015059​

0.02454308​

6.28302760​

3.14151380​

0.99992470​

0.01227154​

0.00007530​

0.02454354​

6.28314588​

3.1415729404​

9​

512​

0.703125​

−162.9726164132​

0.00613600015762483589​

162.9726164132​

0.99996235​

0.00613577​

0.00003765​

0.01227177​

6.28314588​

3.14157294​

0.99998118​

0.00613588​

0.00001882​

0.01227183​

6.28317545​

3.1415877253​

10​

1024​

0.3515625​

−325.9483007953​

0.00306797120144551311​

325.9483007953​

0.99999059​

0.00306794​

0.00000941​

0.00613591​

6.28317545​

3.14158773​

0.99999529​

0.00306796​

0.00000471​

0.00613592​

6.28318284​

3.1415914215​

11​

2048​

0.17578125​

−651.8981355542​

0.00153398199114320724​

651.8981355542​

0.99999765​

0.00153398​

0.00000235​

0.00306796​

6.28318284​

3.14159142​

0.99999882​

0.00153398​

0.00000118​

0.00306796​

6.28318469​

3.1415923457​

12​

4096​

0.08789063​

−1303.7970381577​

0.00076699054433582181​

1303.7970381577​

0.99999941​

0.00076699​

0.00000059​

0.00153398​

6.28318469​

3.14159235​

0.99999971​

0.00076699​

0.00000029​

0.00153398​

6.28318515​

3.1415925766​

13​

8192​

0.04394531​

−2607.5944588431​

0.00038349521591009354​

2607.5944588431​

0.99999985​

0.00038350​

0.00000015​

0.00076699​

6.28318516​

3.14159258​

0.99999993​

0.00038350​

0.00000007​

0.00076699​

6.28318527​

3.1415926355​

14​

16384​

0.02197266​

−5215.1891161110​

0.00019174760065953264​

5215.1891161110​

0.99999996​

0.00019175​

0.00000004​

0.00038350​

6.28318526​

3.14159263​

0.99999998​

0.00019175​

0.00000002​

0.00038350​

6.28318529​

3.1415926459​

15​

32768​

0.01098633​

−10430.3782688848​

0.00009587379999277001​

10430.3782688848​

0.99999999​

0.00009587​

0.00000001​

0.00019175​

6.28318533​

3.14159266​

1.00000000​

0.00009587​

0.00000000​

0.00019175​

6.28318533​

3.1415926673​

16​

65536​

0.00549316​

−20860.7563741179​

0.00004793690037244807​

20860.7563741179​

1.00000000​

0.00004794​

0.00000000​

0.00009587​

6.28318540​

3.14159270​

1.00000000​

0.00004794​

0.00000000​

0.00009587​

6.28318540​

3.1415927001​

17​

131072​

0.00274658​

−41721.5152350710​

0.00002396844875757058​

41721.5152350710​

1.00000000​

0.00002397​

0.00000000​

0.00004794​

6.28318503​

3.14159251​

1.00000000​

0.00002397​

0.00000000​

0.00004794​

6.28318503​

3.1415925149​

18​

262144​

0.00137329​

−83443.0739457073​

0.00001198421813475682​

83443.0739457073​

1.00000000​

0.00001198​

0.00000000​

0.00002397​

6.28318176​

3.14159088​

1.00000000​

0.00001198​

0.00000000​

0.00002397​

6.28318176​

3.1415908785​

19​

524288​

0.00068665​

−166886.0609762367​

0.00000599211218810175​

166886.0609762367​

1.00000000​

0.00000599​

0.00000000​

0.00001198​

6.28318503​

3.14159251​

1.00000000​

0.00000599​

0.00000000​

0.00001198​

6.28318503​

3.1415925148​

20​

1048576​

0.00034332​

−333771.2637536830​

0.00000299606379756521​

333771.2637536830​

1.00000000​

0.00000300​

0.00000000​

0.00000599​

6.28320119​

3.14160059​

1.00000000​

0.00000300​

0.00000000​

0.00000599​

6.28320119​

3.1416005926​

21​

2097152​

0.00017166​

−667544.2439139351​

0.00000149802804700527​

667544.2439139351​

1.00000000​

0.00000150​

0.00000000​

0.00000300​

6.28318503​

3.14159251​

1.00000000​

0.00000150​

0.00000000​

0.00000300​

6.28318503​

3.1415925148​

22​

4194304​

8.583E−05​

−1335151.1090987341​

0.00000074897889323930​

1335151.1090987341​

1.00000000​

0.00000075​

0.00000000​

0.00000150​

6.28289034​

3.14144517​

1.00000000​

0.00000075​

0.00000000​

0.00000150​

6.28289034​

3.1414451678​

23​

8388608​

4.292E−05​

−2670705.5142517439​

0.00000037443289597587​

2670705.5142517439​

1.00000000​

0.00000037​

0.00000000​

0.00000075​

6.28194157​

3.14097079​

1.00000000​

0.00000037​

0.00000000​

0.00000075​

6.28194157​

3.1409707866​

24​

16777216​

2.146E−05​

−5353320.1612446234​

0.00000018679996149670​

5353320.1612446234​

1.00000000​

0.00000019​

0.00000000​

0.00000037​

6.26796661​

3.13398330​

1.00000000​

0.00000019​

0.00000000​

0.00000037​

6.26796661​

3.1339833028​

25​

33554432​

1.073E−05​

−10785541.4998633806​

0.00000009271671709878​

10785541.4998633806​

1.00000000​

0.00000009​

0.00000000​

0.00000019​

6.22211356​

3.11105678​

1.00000000​

0.00000009​

0.00000000​

0.00000019​

6.22211356​

3.1110567792​

26​

67108864​

5.364E−06​

−21976788.0303729475​

0.00000004550255472355​

21976788.0303729475​

1.00000000​

0.00000005​

0.00000000​

0.00000009​

6.10724951​

3.05362476​

1.00000000​

0.00000005​

0.00000000​

0.00000009​

6.10724951​

3.0536247566​

27​

134217728​

2.682E−06​

−51231322.1243512332​

0.00000001951930886290​

51231322.1243512332​

1.00000000​

0.00000002​

0.00000000​

0.00000004​

5.23967458​

2.61983729​

1.00000000​

0.00000002​

0.00000000​

0.00000004​

5.23967458​

2.6198372877​

28​

268435456​

1.341E−06​

−87907152.1214918643​

0.00000001137563868089​

87907152.1214918643​

1.00000000​

0.00000001​

0.00000000​

0.00000002​

6.10724951​

3.05362476​

1.00000000​

0.00000001​

0.00000000​

0.00000002​

6.10724951​

3.0536247566​

 

[fresh_42]

Here is a quick and dirty estimation without trigonometric functions.

The red arch between $A$ and $B$ is of length $\pi/6.$ The line $\overline{BC}$ is $1/2$ which gives an estimation of $\pi \approx 3.$ The blue line $\overline{AB}$ is of length $\sqrt{2-\sqrt{3}}$ which could be estimated for centuries. That gives an estimation of $\pi\approx 3.106.$

So even some scribblings in ancient Greek sand yield some easy approximations. And Archimedes had $3.14$ $250$ BC.

 

[pbuk]

Have neither seen an estimation nor derivation of pi that does not use trig functions.

Have you looked on Wikipedia?

after the 15th iteration , the estimate for pi becomes erratic.

I can't quite follow your description, but I assume it is similar to the method described in Wikipedia. The main problem lies in the calculation $y^2 + u^2$ when y becomes much larger than u. Excel only calculates to about 16 digits of precision so when you are squaring things and adding them together you will only be able to get about 8 digits, after that things will start to go wrong.

 

[mcastillo356]

And the polynomial approach?

 

[mfb]

But if do this, then after the 15th iteration , the estimate for pi becomes erratic.

You run into rounding errors. There are calculation methods that can keep rounding errors small, or you can just use more digits than your spreadsheet tool does.

There are many ways to find approximations of pi. The digit records are all set via series expansions. As an example, pi = 4 (1 - 1/3 + 1/5 - 1/7 +- ...). No trigonometry involved.

 

[malawi_glenn]

No trigonometry involved.

Don't you need to know the arctan function?

 

[malawi_glenn]

@mcastillo356 https://crypto.stanford.edu/pbc/notes/pi/glseries.html

 

[mcastillo356]

@drmalawi, Taylor and MacLaurin emerged to avoid the trigonometric functions, face them, but not erase or condemn to obstracism

 

[malawi_glenn]

@drmalawi, Taylor and MacLaurin emerged to avoid the trigonometric functions, face them, but not erase or condemn to obstracism

If you do happen to sit on the original derivation, free from invoking trigonometric functions - I would happy to read it.

 

[mcastillo356]

If you do happen to sit on the original derivation, free from invoking trigonometric functions - I would happy to read it.

I promise to replicate from my textbook, in case nobody does before. I'll wait 24 hours. And then start on it.

 

[fresh_42]

Archimedes used the regular $96$-gon. Approximating its side length by the sine which is "allowed" since the angle $3.75°$ is quite small yields $\pi \approx 3.139\ldots .$ That was only the lazy 21st-century method. Archimedes probably calculated the actual side length of the figure.

And $250$ BC is also Bsine, BMcLaurin, Bcalculs, and Btranscendence.

 

[mcastillo356]

@drmalawi, preliminaries:

Taylor's Formula

The following theorem provides a formula for the error in a Taylor approximation $f(x)\approx{P_{n} (x)}$ similar to that provided for linear approximation (...).

THEOREM 12

Taylor's Theorem

If the $(n+1)$st-order derivative, $f^{(n+1)} (t)$, exists for all $t$ in an interval containing $a$ and $x$, and if $P_{n} (x)$ is the $n$-th order Taylor polynomial for $f$ about $a$, that is,

$P_{n} (x)=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+...+\dfrac{f^{(n)}(a)}{n!}(x-a)^n$.

then the error $E_{n}(x)=f(x)-P_{n}(x)$ in the approximation $f(x)\approx{P_{n}(x)}$ is given by

$E_{n}(x)=\dfrac{f^{(n+1)}(s)}{(n+1)!}(x-a)^{n+1}$

where $s$ is some number between $a$ and $x$. The resulting formula

$f(x)=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+...+\dfrac{f^{(n)}(a)}{n!}(x-a)^{n}+\dfrac{f^{(n+1)}(s)}{(n+1)!}(x-a)^{n+1}$, for some $s$ between $a$ and $x$,

is called Taylor formula with Lagrange remainder; the Lagrange remainder term is the explicit formula given above for $E_{n}(x)$.

(Note that the error term (Lagrange remainder) in Taylor's formula looks like the next term in the Taylor polynomial would look if we continued the Taylor polynomial to include one more term (of degree $n+1$), EXCEPT that the derivative $f^{(n+1)}$ is not evaluated at $a$ but rather at some (generaly unknown) point $s$ between $a$ and $x$. This makes easy to remember Taylor's formula)

Proof: Observe that the case $n=0$ of Taylor's formula, namely,

$f(x)=P_{0}(x)+E_{0}(x)=f(a)+\dfrac{f'(s)}{1!}(x-a)$

is just the Mean-Value Theorem

$\dfrac{f(x)-f(a)}{x-a}=f'(s)$ for some $s$ between $a$ and $x$.

Also note that the case $n=1$ is just the error formula for linearization given in Theorem 11.

We will complete the proof for higher $n$ using mathematical induction.(See the proof of Theorem 2 in section 2.3). Suppose, therefore, that we have proved the case $n=k-1$, where $k\geq{2}$ is an integer. Thus, we are assuming that if $f$ is any function whose $k$th derivative exists on an interval containing $a$ and $x$, then

$E_{k-1}(x)=\dfrac{f^{(k)}(s)}{k!}(x-a)^k$
where $s$ is some number between $a$ and $x$. Let us consider the next higher case: $n=k$. As in the proof of Theorem 11 (previous), we assume $x>a$ (the case $x<a$ is similar) and apply the Generalized Mean-Value Theorem to the functions $E_{k}(t)$ and $(t-a)^{k+1}$ on $[a,x]$. Since $E_{k}(a)=0$, we obtain a number $u$ in $(a,x)$ such that

$\dfrac{E_{k}(x)}{(x-a)^{k+1}}=\dfrac{E_{k}(x)-E_{k}(a)}{(x-a)^{k+1}-(a-a)^{k+1}}=\dfrac{E'_{k}(u)}{(k+1)(u-a)^{k}}$

Now

$E'_{k}(u)=\dfrac{d}{dt}\left(f(t)-f(a)-f'(a)(t-a)-\dfrac{f''(a)}{2!}(t-a)^2-\ldots-\dfrac{f^{(k)}(a)}{k!}(t-a)^{k}\right)\Bigg |_{t=u}$

$=f'(u)-f'(a)-f''(a)(u-a)-\ldots-\dfrac{f^{(k)}(a)}{(k-1)!}(u-a)^{k-1}$

This last expression is just $E_{k-1}(u)$ for the function $f'$ instead of $f$. By the induction assumption is equal to

$\dfrac{(f')^{(k)}(s)}{k!}(u-a)^{k}=\dfrac{f^{(k+1)}(s)}{k!}(u-a)^{k}$

for some $s$ between $a$ and $u$. Therefore,

$E_{k}(x)=\dfrac{f^{(k+1)}(s)}{(k+1)!}(x-a)^{k+1}$

We have shown that the case $n=k$ of Taylor's Theorem is true if the case $n=k-1$ is true, and the inductive proof is complete

Remark: For any value of $x$ for which $\lim_{x\rightarrow{\infty}}{E_{n}(x)}=0$ we can ensure that the Taylor approximation $f(x)\approx{P_n}$ is as close as we want by choosing $n$ large enough.

Can you figure it out? I am a little bit confused, I really think you are much more capable than me.

Greetings

 

[malawi_glenn]

Can you figure it out?

Figure out what?

I am a little bit confused

Me too. I thought you had Leibniz original derivation of the approximation of π which bear his name, free of usage of trig-functions...

 

[fresh_42]

orig.: http://www.izwtalt.uni-wuppertal.de/Acta/AE1682.pdf#page=57
book: https://books.google.it/books?id=WNUNAQAAIAAJ&pg=PA118#v=onepage&q&f=false
engl.: https://upcommons.upc.edu/bitstream/handle/2117/118726/article10.pdf?sequence=1&isAllowed=y

The original book (first edition) is available for € 740.

 

[MevsEinstein]

I made the following estimations of $\pi$ using elliptic curves and brute force:

$\sqrt{10}$ (probably known already)
third root of 31
fourth root of 97
fifth root of 306
sixth root of 961
...

Here's how I came up with the approximations: https://www.desmos.com/calculator/yokgiknjuj

I also came up with approximations of $\pi$ using $e$: https://www.desmos.com/calculator/yhrpeoqhlw

I can make more approximations if you want :). Maybe using the golden ratio.

 

[fresh_42]

Now, THAT is a URL:
https://3.141592653589793238462643383279502884197169399375105820974944592.eu/

And whoever is interested in it ...
https://3.1415926535897932384626433.../wp-content/uploads/2020/03/pi-314-inhalt.pdf

 

[MevsEinstein]

And whoever is interested in it ...
https://3.1415926535897932384626433.../wp-content/uploads/2020/03/pi-314-inhalt.pdf

If I did my calculations correctly, there are 5,000 digits in each page (except the one with a three on it). There is an indian man who memorized over seventy thousand digits of pi, which is like memorizing fourteen pages of the book.

 

[fresh_42]

If I did my calculations correctly, there are 5,000 digits in each page (except the one with a three on it). There is an indian man who memorized over seventy thousand digits of pi, which is like memorizing fourteen pages of the book.

From the website:

Memorizing the first 100 digits of pi is enough to be accepted into the club/association of friends of the number pi.

 

[WWGD]

I'm pretty sure you can find approximations using Fourier Series .

My heuristics to remember the first digits:

I made up that my Great^n father was born on March 14, 1592, at 6:53 PM:
3 14 1592 653...

But I need an infinite story for the full expansion.

 

[MevsEinstein]

My best approximation of pi so far is $(2903677)^{\frac{1}{13}}$. It is correct to 8 decimal digits.

 

[berkeman]

My best approximation of pi so far is $(2903677)^{\frac{1}{12}}$. It is correct to 8 decimal digits.

3.45604E0 ?

 

[MevsEinstein]

3.45604E0 ?

Oops I meant 13th root.

 

[berkeman]

Oops I meant 13th root.

Do you work for NASA by any chance?

 

[MevsEinstein]

Do you work for NASA by any chance?

No I am still 13

 

[MevsEinstein]

$\pi=\sqrt{6\displaystyle\sum^\infty_{n=1} \frac{1}{n^2}}$. So if we calculate the first terms in the sum, we can get an approximation of pi. But it will take a ton of terms just to get the second decimal digit.

 

[MevsEinstein]

I made another approximation of pi: $(9122171)^{\frac{1}{14}}$

 

[WWGD]

No I am still 13

Actual age, or just an approximation? ;).

 

[MevsEinstein]

Actual age, or just an approximation? ;).

It's an approximation

 

[MidgetDwarf]

is there a particular book you learned series from? you seem to have good intuition with them.

 

[MevsEinstein]

is there a particular book you learned series from? you seem to have good intuition with them.

It was part internet part this book: https://www.amazon.com/dp/0395977258/?tag=pfamazon01-20 I borrowed this book from my seventh grade math teacher and she let me keep it until the end of the year. But I learned about the Basel problem series from 3blue1brown.

 

[Hornbein]

Here's the whole history of the approximation of pi.



Or how about Kate Bush singing the digits of pi?



There's also a weird art movie called pi that I liked.

I had the idea of using pi as a musical sort of time signature. The first measure has three beats, the second one beat, the third four beats, and so forth. But I didn't follow through with this.

 

[Jodo]

It was part internet part this book: https://www.amazon.com/dp/0395977258/?tag=pfamazon01-20 I borrowed this book from my seventh grade math teacher and she let me keep it until the end of the year. But I learned about the Basel problem series from 3blue1brown.

Grant from 3blue1brown is most awesome. At 13 years of age you are doing quite well! Carry on with Grant. Great job

 

[Bystander]

https://www.amazon.com/dp/0312381859/?tag=pfamazon01-20

 

[mathwonk]

i want to argue that the goal is approximating pi without making use of “trig functions” is in some sense hopeless. Indeed these functions are more naturally called circular functions, and to me this is like asking for the length of a semicircle without using properties of the circle. Indeed the sine function evaluated at t is just the y coordinate of the end point of the counterclockwise arc of length t starting at (1,0), Hence arcsin(y) is by definition the length of the counterclockwise arc from (1,0) to the point of the unit circle with second coordinate y. Since thus arcsin is precisely the arclength function, asking for the value of π is the same thing as asking for the functional value 2arcsin(1), (or arccos(-1)).

The OP made a valid point in observing that it seems problematic to try to evaluate trig functions at rational multiples of π when one does not know in advance how to approximate π. But this is beside the point here, since it is the inverse trig functions which are used, and it is precisely because one often obtains rational outputs from trig functions whose inputs are rational multiples of π, that conversely, one obtains rational multiples of π as outputs of inverse trig functions whose inputs are rational numbers. I.e. using inverse trig functions, one does not need to know approximations to π, rather one plugs ordinary rational numbers into power series, such as that named after Leibniz, (but discovered earlier it seems by Madhava), and the partial sums of the outputs give approximations to π.

On the other hand, if one uses the method of Archimedes, approximating the length of the semi circle by the lengths of tangent segments, one is indeed using ordinary trig functions, not their inverses, evaluated at rational multiples of π, but these inputs are done by geometry, i.e. by subdividing a semicircle into an integer number of parts, and then evaluating by actual measurement. This method uses the fact that the function tan(πx)/x converges to π as x—>0.

I.e. rather than trying to plug πx into a power series for tan, we subdivide a unit radius semicircle into n equal parts, draw a right triangle with vertex at the center of the unit circle, right angle vertex at (1,0), and acute angle = π/n at the center (0,0), hence with hypotenuse dividing the semi circle by n, then measure the vertical side opposite the angle π/n, and approximate π by n times this length. I.e. this side opposite the angle of π/n has length tan(π/n), and the approximating polygon to the upper semicircle has length n.tan(π/n), which approaches π as n approaches infinity.

So I claim all approaches to approximating π must involve trig functions one way or another, even series like π^2/6 = SUM (1/n^2), as Euler explains in his wonderful little book “On the analysis of infinities”, chapter X, at least if you notice that (e^t - e^-t)/2 = sinh(t) = i sin(t/i), where i = sqrt (-1). Euler also states in chapter VIII that he himself approximated π to hundreds of digits, using the series for arctan evaluated at 1/sqrt(3), "with incredible labor", and then explains how to greatly reduce this labor by using the addition formula for tan.here is a nice explanation of how Archimedes method can be done in practice, without physical measurement, by using "double angle" trig formulas:
https://arxiv.org/pdf/2008.07995.pdf

 

[fresh_42]

$\pi$ is defined as half the circumference of the unit circle. The circle is parameterized by $t \longmapsto (\cos(t),\sin(t)).$ With that point of view, you cannot even define $\pi$ without sine and cosine.

We can approximate $\pi$ by regular polygons as Archimedes did. This requires some calculations with triangles. Now, sine and cosine are relations in right triangles, so they are hidden there again. However, I would not say that those approximations involve trigonometric functions, only because both are related via triangles. Suited regular polygons allow approximations that do not require the computation of sine values despite the fact that their edge length is basically a sine value. We can get relevant lengths with triangle computations and square roots alone.