Different ways to formally define trigonometric functions

Question

When I first learnt trigonometric functions I was in highschool and obviously the explanation they gave me was mostly intuitive. Now that I have taken my first curse of calculus I learnt a formal definition by using integrals and the notion of sectors in a unit circle. From Calculus by Michael Spivak:

1.-$\pi:=2\int_{-1}^{1}\sqrt{1-x^2}dx$

2.-$A(x):=\frac{x\sqrt{1-x^2}}{2}+\int_x^1\sqrt{1-t^2}$

3.-If $0\leq x\leq \pi$, then $\cos(x)$ is the unique number in $[-1,1]$ such that $A(\cos x)=\frac{x}{2}$.

4.- $\sin x=\sqrt{1-\cos ^2 x}$

Now, the motivation behind this approach of course is very good explaind in the book. The thing is that I was expecting a rigorous treatment based on the same intuitive ideas that they gave me in highschool, meaning by using lengths instead of area. So, what I'm looking for is different approaches other than the one given by professor Spivak and bibliography with the same level of rigor.

Answer 1

Define the unit circle to be the locus $x^2+y^2=1$.

Define $(\cos \theta,\sin \theta)$ to be the point at arc-length $\theta$ from the point $(1,0)$ on the unit circle (measuring counter-clockwise).

So the red segment has length $\theta$ units.

Then we define $(\cos\theta,\sin\theta)$ to be the x- and y-coordinates of the point $D$. Since $D$ is on the unit circle, it immediately follows that $x^2+y^2=\cos^2\theta+\sin^2\theta=1$. It can also be shown geometrically that $$\sin(a+b)=\sin a\cos b+\sin b\cos a$$ From this all other standard identities can be easily derived, and the derivatives of $\sin x$ and $\cos x$ may be proved.

Note: the number $\pi$ can be defined as half the circumference of the unit circle. Then an angle of $2\pi$ radians goes around the whole circle exactly once. From the derivatives mentioned above and this definition of $\pi$ (which implies that $\sin^{-1}(1)=\pi/2$, among other things) we can use some rules of integration to show that $$\text{area of unit circle}=2\int_{-1}^1{\sqrt{1-x^2}\,\mathrm{d}x}=\pi$$

It is important for you to realize that Spivak's are not the original definitions of $\sin x$ and $\cos x$. They are theorems, originally proved based on the definitions given above (also note: the "unit circle" definition and the "right triangle" definition are easily shown to be the same), and now turned into definitions with the "definition" becoming a "theorem."

The reason for using this unusual approach is to simplify algebraic derivations and proofs which would otherwise be more challenging, and to streamline the process of building up the basic facts regarding trigonometric theorems. Whether this is "better" or "worse" is for you to decide, but the straightforward (albeit slightly more tedious) approach given in this post has always been more intuitive and appealing to me.

Answer 2

We can start with remarkably simple differential equations to define two functions: $\sin$ and $\exp$. Then we can use calculus to show how those functions are related in the form of the unit circle.

$\sin 0 = 0$; $\sin^\prime 0 = 1$; $\sin^{\prime\prime} = -sin$

$\exp 0 = 1$; $\exp^\prime = \exp$

These equations represent some very nice intuitions. $\sin$ represents a harmonic oscillator (the farther it is from 0, the faster it accelerates toward 0); and $\exp$ represents exponential growth (the bigger it is, the faster it grows).

The recursion in these definitions allows us to use the fact that $f=g \Rightarrow f^\prime = g^\prime$ to generate a sequence of iterated derivatives for each function:

$\sin, \sin^\prime, -\sin, -\sin^\prime, \sin, \sin^\prime, -\sin, -\sin^\prime, \dots$

$\exp, \exp, \dots$

Since the starting values are given for $\sin 0$, $\sin^\prime 0$, and $\exp 0$, these sequences give us the Maclaurin series for their respective functions, allowing us to calculate the functions to an arbitrary precision.

But let's go deeper. Let's define $E$ as $\theta \mapsto \exp (i\theta)$ and $S$ as $\theta \mapsto \sin^\prime \theta + i \sin \theta$. With a little algebra and some application of the chain rule, we find that the iterated derivatives for these functions are:

$E, iE, -E, -iE, E, iE, -E, -iE, \dots$

$S, iS, -S, -iS, S, iS, -S, -iS, \dots$

And $E(0) = S(0) = 1$.

Since the sequences and starting conditions are the same, that means the functions are the same. That is, $\exp (i\theta) = \sin^\prime \theta + i \sin \theta$, which is Euler's formula.

So for a complex number $x+yi$ with magnitude $1$, its angle $\theta$ is related to its coordinates $(x, y)$ by the equation $(x, y) = (\sin^\prime \theta, \sin \theta)$.

There's an even more intuitive reason why this makes sense.

A point orbiting the origin at a distance of $r$ is a point whose:

• position vector $\mathbf{d}$ of magnitude $r$ points from the origin to the orbiting point

• velocity vector $\mathbf{v}$ of magnitude $r$ points perpendicular to the position vector

• acceleration $\mathbf{a}$ vector of magnitude $r$ points opposite to the position vector

And we can directly apply our definition of $\sin$ from above to show that a point at position $(x, y) = (\sin^\prime t, \sin t)$ at time $t$ indeed has those properties:

• $\mathbf{d} = (x, y)$

• $\mathbf{v} = \frac{d\mathbf{d}}{dt} = (\sin^{\prime\prime} t, \sin^\prime t) = (-\sin t, \sin^\prime t) = (-y, x)$

• $\mathbf{a} = \frac{d\mathbf{v}}{dt} = (-\sin^\prime t, \sin^{\prime\prime} t) = (-\sin^\prime t, -\sin t) = (-x, -y) = -\mathbf{d}$

And according to the starting conditions, $\mathbf{d}_0 = (1, 0)$; $\mathbf{v}_0 = (0, 1)$; $\mathbf{a}_0 = (-1, 0)$

Answer 3

Using

$$x=\sqrt{1-y^2}$$

we have, for $y\in[-1,1]$

$$S(y):=\int_0^y ds=\int_0^y\sqrt{dx^2+dy^2}=\int_0^y \sqrt{1+\frac{y^2}{1-y^2}}dy=\int_0^y \frac{dy}{\sqrt{1-y^2}}$$

and we can define

$$\sin(t):=S^{-1}(t).$$

Note that $ds^2=dx^2+dy^2$ is a version of Pythagora's.