Cover the n-sphere with sub-hemispherical caps

Question

Original Question (answered):

Define a cap (x,Phi) to be the set of all points of the sphere that are within an angle Phi of the point x. $ 0 \le \phi < \frac{\pi}{2} $. (define the angle between 2 points as the angle between these points considered as vectors)

How many such caps do you need to cover the surface of the n-sphere?

Answer: n+2

Expanded Question: What if we bound the angle $\phi$ by a smaller constant, i.e. $ 0 \le \phi < \theta < \frac{\pi}{2} $, for some $\theta$?

Answer 1

Take the intersection of $x_0^2+...+x_n^2=n^2+n$ and $x_0+...+x_n=0$. This is an $n-1$-dimensional surface of an $n$-dimensional sphere.
It contains the points
$(n,-1,-1,-1,...,-1)$, $(-1,n,-1,...,-1)$ and so on.
For any point on the sphere, at least one coordinate is positive, so at least one of the dot products with these points is positive.
So $n+1$ points will do for the $n-1$-dimensional surface.