Why is $\cos(x)dx$ a differential form I am trying to understand the concept of the differential
A differential $d$ is a map that sends functions to 1-forms
A preliminary example is
$$d\sin(x) = \cos(x) dx$$
So the operator $d$ sends a function $\sin(x)$ to a differential form $\cos(x)dx$
My question is that while $dx$ is obviously a differential form, but $\cos(x)$ is just a function. 
What makes a function multiplying a differential form a differential form? i.e why would $\cos(x)dx$ be a differential form when the only identifiable differential form is the quantity $dx$

(Just learning all this myself, so if something seems wrong it probably is.) A $k$-form is a function that takes $k$ vectors and returns a scalar. 
$$\cos(x)\,dx$$
is a $1$-form -- its value evaluated on the vector $(a, b, c)$ is
$$a\cdot\cos(a)$$
A one-form, let's say on $\mathbb{R}^3$:
$$f_1(x, y, z)\,dx + f_2(x, y, z)\,dy + f_3(x, y, z)\,dz$$
is called differential if $f_1, f_2, f_3$ are all differentiable. So $\cos(x)\,dx$ is a differential form.
It's true that $dx$ by itself is a differential form, and it's true that $\cos(x)\,dx$ is also a differential form. It just has a scalar in front of it.
