As an example, the function $\cos(\theta)$ represents the ratio of, relative to the angle $\theta$, the adjacent-side and hypotenuse of a right triangle. Strictly speaking, $\theta$ is measured in any units we want – it's the interpretation/definition of the $\cos(\theta)$ function relative to our chosen "units" for $\theta$ that matters. That's why it's just a matter of economics when choosing between radians, degrees, or anything else.
Of course, when finding arcs on a circle, it's convenient to define the angle $\theta$ in terms of the ratio between the radius of a circle and its circumference, $1/2\pi$ (e.g. when you've covered an angle $\pi$ in these units along the circumference of a circle, you've traveled $\pi/(2\pi)=1/2$ of the circumference). That way, calculating arc-lengths becomes simple multiplication by the angle $\theta$ (this obviously isn't true for degrees!).
However, I don't know how to interepret the power (Taylor/Maclaurin) series for trig functions like $\cos(\theta)$ and $\sin(\theta)$.
$$\cos(\theta)=1-\frac{\theta^2}{2}+\cdots,\,\,\,\,\sin(\theta)=\theta-\frac{\theta^3}{6}+\cdots$$
Why must we use radians in the above series representation? Why don't we use for the "units" of $\theta$, for example, the fraction of the entire circle that it covers (e.g. $1/4$ instead of $\pi/2$, $1$ instead of $2\pi$, etc.)? That would seem more natural and "unitless".
Best Answer
Think of radians as having a special identity multiplication property.
$radians\cdot radians=radians$. The 'unit' is unaffected unlike other units.
In fact, I learned in physics the other day that $radians\cdot meters=meters$, which means that 'radians' aren't even really units.
This allows the Taylor series for trig functions to make sense in terms of units.
Since degrees have the same property, it is simply how we derive the Taylor series of our trigonometric functions that determines that we use radians over degrees.
It has to do with derivatives, which makes a well defined Taylor series for our trig functions if we use radians.
If we use degrees, then the Taylor series for the trig functions are
$$\cos(\theta)=1-\frac{\left(\frac{90}{\pi}\right)^2\theta^2}{3!}+\frac{\left(\frac{90}{\pi}\right)^4\theta^4}{5!}-\dots$$
Of course, this makes sense and all, but the reason it is like this is because $\frac d{d\theta}\cos(\theta)=-\frac{90}{\pi}\sin(\theta)$, if $\theta$ is given in degrees. This results from simple chain rule, which you will learn about in Calculus I.