Why is it important to extend the trigonometric functions to all angles? The title saids it all. I understand the process to evaluate a trig function for any angle but I don't get why it is actually important...
 A: Aside from applications, here are a couple of reasons:


*

*Addition formulas, half-angle, formulas, etc. in trigonometry would have to have a bunch of separate cases to enforce their parameters being on $[0, 2\pi]$. Periodicity simplifies the situation, and there's no reason not to do so.

*For most definitions of the trig functions (Taylor series, differential equations, the connection with $e^{iz}$, etc.), there's no need to restrict the definition to $[0, 2\pi]$; the function in question makes sense or converges on a much larger domain. For that matter, the geometric interpretation corresponding to an angle in the plane makes sense over that larger region.

*Defining trig functions on the entire real line eliminates some techincal annoyances with respect to continuity, differentiability, etc. that would occur if they were defined on a closed interval with a boundary.

*It's generally useful to think of these functions as defined on the circle $S^1 = \mathbb{R}/\mathbb{Z}$ (or the equivalent), and that naturally leads to them being defined on $\mathbb{R}$ and periodic.
A: Well, if you want to evaluate a trig function at some arbitrary angle, but you haven't extended the trig functions there, you have a big problem!
A simple example of an application of such a thing is that the points on the unit circle can be given by $(\cos \theta, \sin \theta)$, where $\theta$ is the* counter-clockwise angular displacement of the point from the positive $x$ axis.
On a closely related note, the trig functions are extremely useful examples of periodic functions.
*: "the" isn't really accurate, because the same angular position can be described by many different angular displacements: e.g. both $0$ and $2\pi$ radians describe the same angular position.
