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 take on 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:
$$\lim_{\theta \rightarrow \pi} \cos(\theta) = -1$$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 VI: 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. $\square$
Something interesting happens if we try to invert the circular functions. Since they are not at all injective, even without the periodicity, we will need to restrict the domain before inverting. It's not exactly clear at this time which domain we should pick. Even once this is done, it isn't clear how one would use the original definitions of the circular functions to obtain an explicit formula for the inverse functions. Fortunately, we can make this easy by using the geometry of the circle.
From the way we defined sine and cosine, we would expect their inverses functions (Which we will denote using an arc- prefix, for arc length) do something to the effect of:
$$x = \cos(\theta) \Rightarrow \arccos(x) = \theta$$ $$y = \sin(\theta) \Rightarrow \arcsin(y) = \theta$$In other words, give the inverse cosine function an x-coordinate and it'll give you an arc length. Give the inverse sine function the corresponding y coordinate on the unit circle, and it'll give you the exact same arc length. In the picture below, the blue arc length $\theta$ is both $\arccos (x)$ and $\arcsin (y)$.
Since we are dealing directly with x and y coordinates, the Cartesian representation of the unit circle will be the most helpful here. We can rearrange it to get $y$ as a function of $x$, which gives us the top half of the unit circle, and similarly we can rearrange it to get $x$ as a function of $y$, giving us the right half. Since these two functions are basically the same, the integrand we'll want for the arc length integrals is the same for both the arcsine and arccosine functions.
$$x^2 + y^2 = 1 \Rightarrow y = \sqrt{1 - x^2} \Rightarrow x = \sqrt{1 - y^2}$$The only thing that will differ in both integrals are the bounds of integration. We see from the picture that we're gonna want to integrate from $0$ to $y$ to get the arcsine, and from $x$ to $1$ to get the arccosine. We therefore get the following integral representations for the inverse circular functions, from which finding the derivatives is straightforward:
$$\arcsin(y) = \int_{0}^{y} \frac{du}{\sqrt{1 - u^2}} \Rightarrow \frac{d}{dy}\arcsin(y) = \frac{1}{\sqrt{1 - y^2}}$$ $$\arccos(x) = \int_{x}^{1} \frac{du}{\sqrt{1 - u^2}} \Rightarrow \frac{d}{dx}\arccos(x) = -\frac{1}{\sqrt{1 - x^2}}$$
Observe that both of these functions have a domain of $[-1, 1]$, but the images are different. The arcsine function counts quadrant I's arc as positive, and quadrant IV's as negative, so the image is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. On the other hand, the arccosine function always counts arc lengths in both quadrant I and quadrant II as positive, so the image is $[0, \pi]$. It is worth repeating for emphasis that from the perspective of arcsine and arccosine as inverses of sine and cosine, both functions measure arc lengths in different quadrants, as seen above.
If we instead choose to picture the arcsine and arccosine functions in isolation, considering only the integral representations, and imagine that their inputs are both the same x coordinate, the picture becomes much more straightforward, as seen below, and it becomes easy to see that the following identity holds:
$$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$$