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Iv always wondered how a sine is computed and related fundamentally to geometry and how it is even defined. Iv seen that mostly this is done with things like COORDINC or something, some coordinate based algorithm, and arc lenght is computed with Cartesian coordinates. I want a

COORDINATE FREE = WITHOUT ANY COORDINATES geometric definition incorporating also the differential equation definition. The input of sine does not need to be x coordinate, it could be an angle, a vector or santa clause as long as its defined corrrectly

Basically sine has second derivative equal to its negative, as does the cosine, in addition to cosine being derivative of sine. the solutions to these diff equations are the respective trig series. So how does this even prove sin of pi is 0, what is even the definition of pi, and how does all this realise itself in geometry, and by geometry don't mean coordinate systems which are opposite to geometry. (when someone asks about a geometric definition of something usually this means coordinate free

Matko
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2 Answers2

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I had asked a similar question to this on this site, in that I wanted to formalise the definition of cosine starting from the geometric intuition. Here it is.

In a sense, you do more or less need coordinates to define the sine and cosine functions, because by definition, they are the length of the sides of the right-angled triangle formed by traveling a distance $\theta$ from a starting point. To have all these notions defined in a "coordinate-free" setting (like Hilbert's axioms for geometry) is a bit of overkill.

Take a look at these (partially completed) notes on trigonometry, where I define the cosine and sine from first principles and then in a remark I discuss how the definitions are equivalent to the familiar power series $$\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}\,x^{2k}.$$

Luke Collins
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  • Yet the purely analytic definition of trigonometric functions (as exposed in Bourbaki, for instance) is via power series. In this case, one has to explain how they play a role in geometry. – Bernard Apr 03 '20 at 21:12
  • @Bernard In the notes I linked, I show that the geometric definition is equivalent to this (I do not start from this). – Luke Collins Apr 03 '20 at 21:14
  • it's bizarre that at least at my university iv never seen sine defined properly, iv never seen the explanation of why the series definition agrees with the geometric one except maybe with complex numbers, but that's not quite it. But I would like a purely geometric coordinate free definition if it is possible to do so, no matter how much overkill it would be no coordinate definition means anything to me – Matko Apr 04 '20 at 17:14
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    @ArcDDD Did you see the notes I linked? – Luke Collins Apr 04 '20 at 17:26
  • Do you use coordinates in them? – Matko Apr 04 '20 at 17:37
  • I scrolled quickly, if they are coordinate less than I'll be happy to read them – Matko Apr 04 '20 at 17:38
  • @ArcDDD It is not bizarre at all that you have not seen $\sin$ or $\cos$ defined without coordinates, because modern geometry is all done in $\mathbb R^n$, so that's why I'd say that it is esoteric to try and define sine and cosine without coordinates. Fundamentally, sine and cosine are defined to be the coordinates of a point on the circle $x^2+y^2=1$ in the plane. – Luke Collins Apr 04 '20 at 19:00
  • Esoteric? Lol, I didn't say it was bizzare I haen seen a coordinate free definition of sine, but they I haven seen any that relates both infinite series and geometry.. – Matko Apr 04 '20 at 20:30
  • Why did u use that word esoteric makes no sense – Matko Apr 04 '20 at 20:31
  • You can find a relationship between the geometry and power series in the notes I linked. – Luke Collins Apr 04 '20 at 20:47
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Here's one approach. First, using coordinates, define arc length $\theta $ on the unit circle as the limit of the sum of the length of inscribed chords. Prove that for any $0 \le \theta < \text { circumference of circle }$ there is exactly one $(x,y)$ on the unit circle such that the arc length from (1,0) to (x,y) is $\theta .$ Define $\sin \theta =y, \cos \theta=x.$ Then prove that $$\lim_{\theta \to 0}\frac {\sin \theta}{\theta}=1$$ and thence prove that the derivatives of $\sin \theta$ and $\cos \theta$ are $\cos \theta$ and $-\sin \theta$ respectively, from which it follows that the $\sin \theta$ function is a solution of $f''(\theta)+f(\theta)=0, f(\theta)=0,f'(\theta)=1.$. $$*$$ Second, define as power series $$\text {ssin}\theta=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-...$$ and $$\text {ccos}\theta=1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-...$$ Thence prove that the derivatives of $\text { ssin } \theta \text { and } \text {ccos} \theta \text { are ccos } \theta \text { and -ssin } \theta$ respectively,from which it follows that the $\text { ssin }\theta$ function is also a solution of $f''(\theta)+f(\theta)=0, f(\theta)=0,f'(\theta)=1.$. $$*$$ Third, prove that the solution of $f''(\theta)+f(\theta)=0, f(\theta)=0,f'(\theta)=1$ is unique. Thus the geometrically defined sine function equals its power series expansion. The same could and should be done for the cosine function.

P. Lawrence
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