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 $dx$

Differential forms can be added, multiplied by scalars, and given a product, namely the wedge product. Let $E^k (M)$ denote the set of all smooth $k$-forms on $M$, a manifold. 

Proposition: $E^0 (M)$, the set of all smooth 0-forms, can be identified with $C^\infty (M)$. 

Forms can be given a multiplication, namely the wedge product. In your case, you have $\cos (x) \in E^0 (M)$ and $dx \in E^1 (M)$. Writing $\cos (x) dx$ is actually shorthand for writing $\cos (x) \wedge dx \in E^{0 + 1} (M)$ and hence is a 1-form. 
