Approach #1 : We could simply use Scipy's cdist with its cosine distance functionality -
from scipy.spatial.distance import cdist
val_out = 1 - cdist(array1, array2, 'cosine')
Approach #2 : Another approach using matrix-multiplication -
def cosine_vectorized(array1, array2):
    sumyy = (array2**2).sum(1)
    sumxx = (array1**2).sum(1, keepdims=1)
    sumxy = array1.dot(array2.T)
    return (sumxy/np.sqrt(sumxx))/np.sqrt(sumyy)
Approach #3 : Using np.einsum to compute the self squared summations for another one -
def cosine_vectorized_v2(array1, array2):
    sumyy = np.einsum('ij,ij->i',array2,array2)
    sumxx = np.einsum('ij,ij->i',array1,array1)[:,None]
    sumxy = array1.dot(array2.T)
    return (sumxy/np.sqrt(sumxx))/np.sqrt(sumyy)
Approach #4 : Bringing in numexpr module to offload the square-root computations for another method -
import numexpr as ne
def cosine_vectorized_v3(array1, array2):
    sumyy = np.einsum('ij,ij->i',array2,array2)
    sumxx = np.einsum('ij,ij->i',array1,array1)[:,None]
    sumxy = array1.dot(array2.T)
    sqrt_sumxx = ne.evaluate('sqrt(sumxx)')
    sqrt_sumyy = ne.evaluate('sqrt(sumyy)')
    return ne.evaluate('(sumxy/sqrt_sumxx)/sqrt_sumyy')
Runtime test
# Using same sizes as stated in the question
In [185]: array1 = np.random.rand(10000,100)
     ...: array2 = np.random.rand(5000,100)
     ...: 
In [194]: %timeit 1 - cdist(array1, array2, 'cosine')
1 loops, best of 3: 366 ms per loop
In [195]: %timeit cosine_vectorized(array1, array2)
1 loops, best of 3: 287 ms per loop
In [196]: %timeit cosine_vectorized_v2(array1, array2)
1 loops, best of 3: 283 ms per loop
In [197]: %timeit cosine_vectorized_v3(array1, array2)
1 loops, best of 3: 217 ms per loop