Why is it allowed to break up $\frac{v\,\cdot\,w}{\|v\|\|w\|}$ as $\frac{v}{\|v\|}\cdot\frac{w}{\|w\|}$, where "$\,\cdot\,$" is the dot product? I see in Strang Linear Algebra (Ed 5) the cosine formula:
$$\frac{v\cdot w}{\left \| v \right \|\left \| w \right \|} = \cos\theta $$
In one of the problem set answer key, a solution implies that:
$$\frac{v}{\left \| v \right \|} \cdot \frac{w}{\left \| w \right \|} = \frac{v\cdot w}{\left \| v \right \|\left \| w \right \|}$$
If it were all simple scalar multiplication, $\frac{ab}{cd}$, I see you can break it up like that, but it wasn't clear to me that I can break up a dot-operation in the same way.
What's the argument that allows me to say this?
 A: For all $c\in \mathbb{R}$ and $v,w\in\mathbb{R}^n$, we have the following two dot product properties:
$$(cv\cdot w)=c(v\cdot w)$$
and
$$v\cdot w=w\cdot v.$$
So in your case, we have
$$\begin{align}
\frac{v}{\|v\|}\cdot\frac{w}{\|w\|}&=\frac{1}{\|v\|}(v\cdot\frac{w}{\|w\|})&\text{(first property)}
\\ & = \frac{1}{\|v\|}(\frac{w}{\|w\|}\cdot v) & \text{(second property)}
\\ & = \frac{1}{\|v\|}\frac{1}{\|w\|}(w\cdot v) & \text{(first property)}
\\ & = \frac{1}{\|v\|}\frac{1}{\|w\|}(v\cdot w) & \text{(second property).}
\end{align}$$
So this gives the desired equality. You probably won't need this level of rigor if this is an introductory linear algebra class, but I am trying to show how you would deduce the result with minimal axioms/rules, which is a relevant skill if you go on to take harder proof-based classes. Also, after having done enough similar calculations, it will probably become much more obvious to you.
A: Well on the numerator it is just the dot product and they are divided by scalars. And the following property holds for dot product. $$\vec A.(\lambda \vec  B) = \lambda (\vec A.\vec B) $$. You can take scalars out of the dot product or take dot product of scaled vectors.
