Triangle equality for cosine similarity in high dimensions I'm trying to understand whether I can use the following equality in my application -- for $u,v,w \in \mathbb{R}^d$:
$$\cos(u,w)\approx \cos(u,v)\cos(v,w)$$
Where $\cos(x,y)$ gives cosine of the angle between vectors $x$,$y$
$$\cos(x,y)=\frac{\langle x, y\rangle}{\|x\| \|y\|}$$
In simulations, I'm finding it becomes a near perfect fit for $d>100$ when $v=f_1(u), w=f_2(v)$ where $f_i(x)$ is a random perturbation of $x$. For instance, $f_i(x)$ could be:

*

*performing random simple rotation of $x$ with angle $\le \frac{\pi}{4}$ (details)

*adding IID standard normal random variable to each entry of $x_i$ (details)

In both cases, this equation gives a good fit even though individual cosines are far from $1$.
Questions:

*

*What is a high-level explanation of this behavior?


*What restrictions on randomly sampled $f_i$ will let me justify using triangle equality for cosine similarity in high dimensions?
$$\cos[u,f_2(f_1((u))]\overset{P}{=} \cos[u,f_1(u)]\cos[f_1(u),f_2(f_1(u))]+O(d^{-\frac{1}{2}})$$
(note, there's a dimension-free bound on approximation error of this formula here, making it applicable in my setting requires turning it into probabilistic bound, and removing "dimension-free" part)
 A: $\newcommand{\R}{\mathbb R}$Here is a straightforward explanation in the case of adding iid standard normal random variables (r.v.'s). Here we have random vectors
\begin{equation}
    U:=X,\quad V:=X+Y,\quad W:=X+Y+Z, 
\end{equation}
where $X,Y,Z$ are independent standard normal random vectors in $\R^d$. Letting $\cdot$ denote the dot product, we have
\begin{equation}
    EU\cdot V=EX\cdot X+EX\cdot Y=d+0=d,
\end{equation}
\begin{equation}
    EV\cdot W=EX\cdot X+EY\cdot Y+2EX\cdot Y+EX\cdot Z+EY\cdot Z \\ 
    =d+d+2\times0+0+0=2d,
\end{equation}
\begin{equation}
    EU\cdot W=EX\cdot X+EX\cdot Y+EX\cdot Z=d+0+0=d.
\end{equation}
Also, since each of the dot products $U\cdot V,V\cdot W,U\cdot W$ is the sum of $d$ iid r.v.'s with finite second moments, we have
\begin{equation}
    Var\,U\cdot V+Var\,V\cdot W+Var\,U\cdot W=Cd
\end{equation}
for some universal real constant $C>0$.
So, by Chebyshev's inequality,
\begin{equation}
    U\cdot V\sim_P d,\quad V\cdot W\sim_P 2d,\quad U\cdot W\sim_P d,  
\end{equation}
where $A\sim_P B$ means that $A/B\to1$ in probability (as $d\to\infty$).
Similarly,
\begin{equation}
    U\cdot U\sim_P d,\quad V\cdot V\sim_P 2d,\quad W\cdot W\sim_P 3d.   
\end{equation}
So,
\begin{equation}
    \cos(U,V)=\frac{U\cdot V}{\sqrt{U\cdot U}\sqrt{V\cdot V}}\sim_P\frac1{\sqrt2}, 
\end{equation}
\begin{equation}
    \cos(V,W)=\frac{V\cdot W}{\sqrt{V\cdot V}\sqrt{W\cdot W}}\sim_P\frac2{\sqrt6}, 
\end{equation}
\begin{equation}
    \cos(U,W)=\frac{U\cdot W}{\sqrt{U\cdot U}\sqrt{W\cdot W}}\sim_P\frac1{\sqrt3}.  
\end{equation}
We conclude that indeed
\begin{equation}
    \cos(U,V)\cos(V,W)\sim_P \cos(U,W). 
\end{equation}

The similar conclusion for other functions $f_i$ is perhaps due to linearization.
