Given a set of $N$ points distributed over a unit ball in $k$ dimensions, if you were to compute the distance of all these points to a single fixed reference point and rank them, would the rankings change if you switched between the Euclidean norm $\sqrt{\sum^k_i (x_i-y_i)^2}$ and the cosine similarity $1-\sum^k_i (x_i \cdot y_i) $?
It's easy to numerically show that the rankings don't change for a given $N, k$,
import numpy as np
N, k = 10**5, 50
# Create N vectors on a k-dimensional sphere
X = np.random.normal(size=(N,k))
X /= np.linalg.norm(X, axis=1).reshape(-1,1)
# Compute differences between the first vectors and all others
dist_cosine = 1 - X.dot(X[0])
dist_euclid = np.sqrt(((X-X[0])**2).sum(axis=1))
# Rank differences
idx_cosine = np.argsort(dist_cosine)
idx_euclid = np.argsort(dist_euclid)
print (idx_cosine == idx_euclid).all() # Always True
Hint: $\sum (x_i - y_i)^2 = \sum x_i^2 + \sum y_i^2 - 2 \sum x_i y_i$. Notice that the first sum is constant because you're on a sphere, and the second is constant because... well, because $y$ is a single vector.