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I want to define trigonometric function (say sine) formally with the definition that the sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse. But there is a problem with defining an angle (and the measure of an angle) without knowing trigonometric functions. Some people say that the measure of an angle is the ratio of the lenth of the arc to the lenght of the radius. But I don't know how to define an arc without trigonometric functions.

The solution may be to define sine and cosine with power series. I know this approach and it's fine, but I'm interested in classical definition.

So I came up with my own idea. Let $x\in (0,90)$ and let $P$ be a model of Hilbert's plane euclidian geometry, $\mu$ is segments measure, $\nu$ is angles measure such that the emasure of the right angle is $90$, and $\triangle abc$ is a right triangle in which $\nu(\angle abc)=x$. Then $\sin x=\frac{\mu(ac)}{\mu(ab)}$.

This definition requires proving many theorems (for instance existance of measures and triangle with given angles) and you have to prove that the definition doesn't depnd on the choice of the model, choice of the segments measure and choice of the right triangle.

The question is: Can we define angles and sine without referring to Hilbert' theory? Maybe it's possible to define measure of angles in euclidian model $\mathbb{R}^2$. I think the key part of the definition must be additivity of the measure, as it is in Hilbert's theory.

Kulisty
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To define an angle it is enough to have a notion of length of a smooth curve or a notion of area enclosed by a smooth, closed, simple curve. If two half-lines $OP, OQ$ share their origin $O$, we may consider a unit circle $\Gamma$ centered at $O$ and consider the circle sector delimited by $OP$ and $OQ$. Then the amplitude of $\widehat{POQ}$ can be defined either in terms of the area of the previous circle sector or the length of its arc. On this set of amplitudes there is a natural equivalence relation given by the fact that a whole turn around a circle brings us in the exact original position.
There also is a delicate point in dealing with $\widehat{POQ}$ and $\widehat{QOP}$ as the same angle or opposite angles, i.e. in using oriented lengths/areas or not.

Jack D'Aurizio
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  • What do you exactly mean by "the circle sector delimited by $OP$ and $OQ$"? I think the arc can be defined as the intersection of halfplanes and circle, but you must prove that it is indeed a curve i.e point a continuous function $f:[a,b]\rightarrow\mathbb{R}^2$ whose image is the arc. – Kulisty Aug 22 '16 at 11:53
  • The circle is a convex subset of the plane with a smooth boundary, hence it is well defined the area of the intersection with two half planes and a circle arc is a curve for sure. For non believers, there is just the canonical parametrization of the boundary in terms of some and cosine. – Jack D'Aurizio Aug 22 '16 at 12:09
  • That's the point. You need sine and cosine to define an arc. So how to define sine and cosine without arcs? I think it's a vicious circle but I may be wrong. – Kulisty Aug 22 '16 at 13:41
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    You may define the elementary trigonometric functions in terms of the complex exponential, and the latter through a series. Kind of unusual but it works, it does not lead to a circular argument (pun intended) – Jack D'Aurizio Aug 22 '16 at 13:46
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    You do not need the sine and cosine to define an arc, nor to define its arc length. Arcs are defined implicitly using the equation $x^2+y^2=1$ or with an explicit parameterization using the functions $y=\pm \sqrt{1-x^2}$ or $x=\pm \sqrt{1-y^2}$. Arc length is an integral of a simple continuous function over a closed segment, a formula for that function being obtained, starting from, say, $y=\sqrt{1-x^2}$, by the arc length integral you learned in multivariable calculus, i.e. $\int \sqrt{dx^2+dy^2}$. – Lee Mosher Aug 24 '16 at 20:41
  • Clearly true, and that also gives a "Jacobian way" for defining the sine function, i.e. as the inverse function of a primitive of $\frac{1}{\sqrt{1-x^2}}$. To do that, obviously, we need the fundamental theorem of calculus (so the notion of area). – Jack D'Aurizio Aug 24 '16 at 20:49
  • I'm not sure about it, but this way I can define the whole arc (for the angle equal to $\pi$ or $2\pi$), but what if I want the arc of an angle $\alpha\in [0,\pi]$? I would need the parametrization $y=\sqrt{1-x^2}$ on the interval $[\cos \alpha,1]$ and hence I need cosine. – Kulisty Aug 29 '16 at 15:44
  • The implicit function theorem ensures that such a parametrization exists, and if you want an explicit parametrization, the complex exponential, as said before, is just fine. – Jack D'Aurizio Aug 29 '16 at 15:49