1. Measuring the Unit Circle
2. Defining the Circular Functions
3. Measuring the Unit Hyperbola
4. Defining the Hyperbolic Functions
5. ...

Our immediate objective before defining the circular functions is to find a way to describe the unit circle analytically. Rather than directly using the traditional $x^2 + y^2 = 1$ which follows immediately from the Pythagorean Theorem and the fact that every point on a circle is the same distance away from the center, we will use a parametrization of the unit circle.

The parametrization works like this: when a chord is made from the point (-1, 0) to any other point on the unit circle, the chord must cross the y-axis at some y-coordinate $t$. The coordinates of the point on the unit circle where the chord goes are then rational functions of $t$.

Proposition I: The unit circle is rationally parametrized by:

$$x = \frac{1 - t^2}{1 + t^2}$$ $$y = \frac{2t}{1 + t^2}$$

Where $t$ is the $y$-coordinate of the point where a chord from the point $(-1, 0)$ to some point on the unit circle crosses the $y$-axis.

Proof: Let O be the origin. Draw the chord from the point P $= (-1, 0)$ to any point C on the circle. Let T be the point with y-coordinate $t$ where this chord PC crosses the y-axis. Drop the altitude CX perpendicular to the $x$-axis. Observe that the triangles PTO and PCX are similar triangles. That is, they form the same slope. This means there must exist some scaling factor $s(t)$ from PTO to PCX. Also observe that PO = 1 since it is a radius of the circle.

The diagram above shows the lengths of triangle PTO before being scaled up to triangle PCX. In particular, note that $1+x = s(t)$ and $y = s(t)\,t$. Applying the Pythagorean Theorem to triangle PCX then gives us:

$$\begin{align*} s(t)^2\,(1+t^2) &= (1+x)^2 + y^2\\ &= x^2 + 2x + 1 + y^2\\ &= 2x + 2\\ &= 2(x + 1)\\ &= 2s(t)\\ \Rightarrow s(t) &= \frac{2}{1+t^2} \end{align*}$$

The result then follows from straightforward manipulation. $\square$

This parametrization can be immediately generalized to all circles by simply multiplying the $x$ and $y$ coordinates by some radius $r$. It is not hard to check that this does in fact give a circle. We only need to check that this new parametrization satisfies $x^2 + y^2 = r^2$.

With a parametrization for the circle under our belt, our next endeavor shall be to find the circumference of the circle and its area. The former is easy once the underlying work of defining integration and arclength has been done. Recall that the arclength of a parametric curve is given by the integral:

$$L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$$

In order to make use of this arclength formula, we first need to take derivatives of each of the functions in the parametrization. Fortunately rational functions don't require any fancy tricks. Despite this, the result of the calculation turns out to be rather interesting.

Proposition II: The derivatives of the functions defining the parametrization of a unit circle are given by:

$$\frac{dx}{dt} = -s(t)\,y(t)$$ $$\frac{dy}{dt} = s(t)\,x(t)$$

Now we are finally ready to prove an important theorem.

Proposition III: The ratio of a circle's circumference to its diameter is always the same.

Proof: Since the parametrization of the unit circle must satisfy:

$$x^2 + y^2 = r^2$$

We have that:

$$\begin{align*} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} &= \sqrt{\frac{1}{r^2} s^2y^2 + \frac{1}{r^2} s^2\,x^2}\\ &= \frac{1}{r} s \sqrt{y^2 + x^2}\\ &= \frac{1}{r} s \sqrt{r^2}\\ &= s \end{align*}$$

Which is what we want to integrate to find the arc length. But what interval are we going to integrate over? The question is equivalent to figuring out which two chords we can slice the circle with that will account for the whole circumference. Unfortunately, there are no such chords, as accounting for the whole circumference would essentially require a tangent line going through $(-r, 0)$. Fortunately, we can get as close to that tangent line as we want by selecting larger and larger values for the y-coordinates our chords will cross. Thus, we will want to integrate over the entire real line.

By using the substitution $ru= t \Rightarrow r\,du = dt$, the circumference calculation becomes:

$$\begin{align*} C &= \int_{-\infty}^{\infty} s\, dt\\ &= 2r \int_{-\infty}^{\infty} \frac{du}{1 + u^2} \end{align*}$$

Which shows that the ratio between the circumference and the diameter of any circle will always come out to be the same integral, as desired. $\square$

The above discovery motivates us to make the following definition:

Definition: The number $\pi$ is the ratio of a circle's circumference to its diameter. This is equivalent to:

$$\pi = \int_{-\infty}^{\infty} \frac{du}{1 + u^2}$$

Furthermore, we also have that the circumference of a circle is given by:

$$C = 2\pi r$$

Proposition IV: The area of a circular sector of radius $r$ subtended by an arc of length $\Theta$ is given by $\frac{1}{2}r \Theta$.

Proof: Let $\Theta$ be the arclength from the point at $t = a$ to $t = b$ for some real $b > a$. Then, by the arclength formula, we have: $$\Theta = \int_{a}^{b} s\, dt$$ We can use the parametric area formula to compute the area of the circular sector over this same interval of values: $$\begin{align*} A &= \frac{1}{2} \int_{a}^{b} x\frac{dy}{dt} - y\frac{dx}{dt} \,dt\\ &= \frac{1}{2} \int_{a}^{b} \frac{1}{r}sx^2 + \frac{1}{r}sy^2\,dt\\ &= \frac{1}{2} \int_{a}^{b} \frac{1}{r} sr^2\, dt\\ &= \frac{1}{2} r \int_{a}^{b} s\, dt\\ &= \frac{1}{2} r\Theta \end{align*}$$

Which is the desired result. $\square$

The formula for the area of the circle then follows as a corollary by integrating over the entire real line.

Exercises

Exercise I: Complete the proof of Proposition I.

Exercise II: Prove Proposition II.

Exercise III: Show that the integral $\int_{-\infty}^{\infty} \frac{du}{1 + u^2}$ is less than or equal to 4.

Intuitively, we define the circular or trigonometric functions as accepting as an input the length of an arc going counterclockwise along the unit circle from the point $(1, 0)$ and giving as an output the x and y coordinates of the point on which that arc ends. The picture of this definition is shown on the left.

This intuitive definition is usually presented without additional comment, which is usually fine for all practical purposes. However, our work in the previous act puts us in a good position to make this intuitive definition much more precise. We will define the circular functions in terms of the rational parametrization and the $\theta$ function:

$$\theta(t) = \int_{0}^{t} \frac{2\, du}{1 + u^2}$$

Which gives us precisely the arclength shown in the unit circle diagram. The benefits of this approach will make themselves clear as we progress in our development of the theory of the circular functions.

Definition: The sine function $\sin: \mathbb{R} \rightarrow [-1, 1]$ and the cosine function $\cos: \mathbb{R} \rightarrow [-1, 1]$ are functions defined to be $2\pi$-periodic and satisfy the functional equations:

$$\sin(\theta(t)) = \frac{2t}{1+t^2}$$ $$\cos(\theta(t)) = \frac{1-t^2}{1+t^2}$$

This definition is conceptually identical to the unit circle definition. However, since it has more moving parts, it's worth stopping to make sure that we didn't screw anything up by putting it this way.

Recall that $\theta(t)$ is invertible, and that its image is the interval $(-\pi, \pi)$. This means that for any real number $r \in (-\pi, \pi)$ there is some corresponding $t_0 \in \mathbb{R}$ such that $\theta(t_0) = r$.

This value $t_0$ may be plugged into the right side of the definitions if $r$ is plugged into one of the circular functions. Thus, the functional equation is a sensible way to define the circular functions over the whole interval $(-\pi, \pi)$.

Since the interval $(-\pi, \pi)$ has a length of $2\pi$, imposing $2\pi$ periodicity does not lead to any conflict as to which values the circular functions have in $(-\pi, \pi)$, and thus the functions are successfully extended to all real numbers. Well, almost. We still haven't defined the functions at inputs which are odd integer multiples of $\pi$. To rectify this, we observe that:

$$\begin{align*} \lim_{\theta \rightarrow \pi} \sin(\theta) &= \lim_{t \rightarrow \infty} \sin(\theta(t))\\ &= \lim_{t \rightarrow \infty} \frac{2t}{1+t^2}\\ &= 0\\ \end{align*}$$

And analogously, that:

$$\begin{align*} \lim_{\theta \rightarrow \pi} \cos(\theta) &= \lim_{t \rightarrow \infty} \cos(\theta(t))\\ &= \lim_{t \rightarrow \infty} \frac{1-t^2}{1+t^2}\\ &= -1\\ \end{align*}$$

So we simply take these to be the values of the circular functions at $\pi$ as part of the definition, as they are the values that keep the functions continuous. Periodicity takes care of all other odd integer multiples of $\pi$. With this final stipulation, the circular functions become defined as desired. We are now ready to prove the following important result:

Proposition V: The sine and cosine functions are differentiable, with derivatives:

$$\frac{d}{d\theta} \sin(\theta) = \cos(\theta)$$ $$\frac{d}{d\theta} \cos(\theta) = -\sin(\theta)$$

Proof: The differentiability of sine and cosine follow from the fact that $x(t)$, $y(t)$, and $\theta(t)$ are all differentiable. Taking derivatives with respect to $t$ on both sides of the definition of sine, we can use the Chain Rule to compute:

$$\begin{align*} \frac{d}{dt}\sin(\theta) &= \frac{dy}{dt}\\ \frac{d}{d\theta}\sin(\theta)\,\frac{d\theta}{dt} &= s(t)\,x(t)\\ \frac{d}{d\theta}\sin(\theta)\,s(t) &= s(t)\,x(t)\\ \frac{d}{d\theta}\sin(\theta) &= x(t)\\ &= \cos(\theta) \end{align*}$$

This gives us the derivative over $(-\pi, \pi)$, and it extends to almost all of the reals by periodicity. The result is extended to odd integer multiples of $\pi$ by a theorem that says any function that is differentiable everywhere around a point is also differentiable at that point with matching derivatives (See Spivak's Calculus, citation will be provided later). The derivative of the cosine function is shown to be negative sine in a similar way. $\square$