So far I've been stuck into this problem. It's easy to get geometrically, but I can't find a rigorous proof using only vectors and the properties of inner product and norms.
Let $v, w \in \mathbb{R^2}$. Using the standard inner product on $\mathbb{R^2}$, prove that if $\theta$ is the angle between $v$ and $v-w$, $\psi$ is the angle between $-w$ and $v-w$, and $\alpha$ is the angle between $v$ and $-w$, then $\alpha = \psi + \theta$ (as shown on my awful picture below).
That is, prove that
$$\arccos\left(\frac{v \cdot (-w)}{|v||-w|}\right) = \arccos\left(\frac{v \cdot (v-w)}{|v||v-w|}\right) + \arccos\left(\frac{-w \cdot (v-w)}{|-w||v-w|}\right)$$
Maybe I'm missing a property from the inverse cosine function. Any tip? Thanks.
You can use:
\begin{equation} \arccos(x)+\arccos(y)=\arccos\left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right) \end{equation} with $x=\frac{v \cdot (v-w)}{|v||v-w|}$ and $y=\frac{-w \cdot (v-w)}{|-w||v-w|}$.
You can get that formula applying a simple law:
if we have \begin{equation} \cos\alpha=x\\\cos\beta=y \end{equation}
then $\alpha+\beta=\arccos x+\arccos y=\arccos[\cos(\alpha+\beta)]$. Now:
\begin{equation} \arccos[\cos(\alpha+\beta)]=\arccos[\cos\alpha\cos\beta-\sin\alpha\sin\beta]=\\=\arccos\left[\cos\alpha\cos\beta-\sqrt{1-\cos^2\alpha}\sqrt{1-\cos^2\beta}\right]=\\=\arccos\left[\cos(\arccos x)\cos(\arccos y)-\sqrt{1-\cos^2\arccos x}\sqrt{1-\cos^2\arccos y}\right]=\\=\arccos\left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right) \end{equation}
with $0\le\alpha,\beta\le\frac{\pi}{2}$ (like in your problem).