I've stumbled over an interesting question. In $\cos(x)$, $x$ is measured in, say, radians. When I expand cosine in Taylor's series, I have the terms with $x^3$, $x^5$ etc. so I am summing up $\text{rad}^3$, $\text{rad}^5$ etc. to get the dimensionless value of cosine. Obviously something is not right here.
I couldn't find any clarification on this issue, so I would appreciate some explanation. I've got one idea myself, but I am quite unsure about it:
If $f = \cos(x)$, $x$ can be expressed as $|x|\cdot u$, where $|x|$ is the value and $u$ the unit of the argument, in our case $u = \text{rad}$. In the Taylor's expansion we have $f^{(n)}(x_0)\cdot(x-x_0)^n$.
$f'(x) = \dfrac{\partial f}{\partial x}$, unit of $\partial\cos$ is 1, since cosine is dimensionless, while the unit of $\partial x$ is radian. Thus the unit of the $n$-th derivative of cosine is $1/\text{rad}^n$.
We then come to each term in the Taylor's series for cosine being dimensionless, thus resulting in the dimensionless sum.
Does it make sense?
EDIT: Since the discussion in the comments has concentrated around radians being dimensionless units, please consider any function with any unit of the argument. The whole consideration remains the same. For example, energy can be expressed as $E=mc^2 = f(m)$. The expansion will then have the terms including $\text{kg}$, $\text{kg}^2$, $\text{kg}^3$ etc. Something should “compensate” these powerful kilograms, and I think it's $f^{(n)}$ that does. But I'd like to hear a confirmation from someone on better terms with mathematics than I am.
Note, that the Taylor series also involves derivatives. And what is a derivative? Let us denote the units of $x$ as $[x] = \text{s}$ and $[f(x)] = \text{m}$ (You can think of time, and space, respectively).
Then a derivative is $f'(x_0) = \lim_{x\rightarrow x_0} \frac{f(x_0)-f(x)}{x_0-x}$. So, plugging in the units, we get $[f'(x)] = \text{m}/\text{s}$. Similarly, $[f''(x)]=\text{m}/\text{s}^2$, and you can continue this onwards. Again, intuitively the derivative is the velocity, and the second derivative is the acceleration, so you can see that if you think of $f$ and $x$ as space/time, the units again check out.
Now, the Taylor expansion looks like this:
$$f(x) = f(x_0) + f'(x_0)(x-x_0) + \frac12f''(x_0)(x-x_0)^2 + ... \approx \text{m} + \frac{\text{m}}{\text{s}} \text{s} + \frac {\text{m}}{\text{s}^2}\text{s}^2 + ... = \text{m}+\text{m}+\text{m} ... = \text{m}$$
(I have used the $\approx$ sign to transfer from precise expression to the measurement units. Hope that this is clear).