What is the distribution of angles for a point uniform on the unit $n$ sphere? Take $x$ uniform on the unit sphere in $n+1$ dimensions. $x$ can be represented as $n$ angles, one from $0$ to $2\pi$ and the others in $[0,\pi)$.
What is the joint distribution of these angles? I know they are not uniformly distributed in this space. (If they were, then caps would be overrepresented) 
Are the angles independent?
 A: This is an old question, but a good one, and one that has some interesting connections across topics that might seem unrelated- the uniform distribution on the n-sphere has a relationship derived in information geometry to the Jeffreys prior for multinomial probabilities- taking the square root of probabilities results in a diffeomorphism between the unit simplex and the unit sphere, and the Jeffreys' prior after transformation is uniform.
The easiest way to approach this is to start with a uniform probability density in Euclidean space, and then to apply a transformation to get the spherical coordinates- which, through the Jacobian of the inverse transformation, gives us the pdf in spherical coordinates. If X is uniform on the n-sphere (n+1 coordinates but with a restricted length), the uniform distribution is proportional to the reciprocal of n-volume of that sphere. Since this is constant, and the radius of the sphere is unimportant to the angular coordinates, I'll restrict my attention to the unit n-sphere and drop the normalization constant (though I will come back to that). For now, the relevant information is that the density is constant.
The conversion from Cartesian coordinates to spherical is straightforward, and the Wikipedia article on the n-sphere gives a good overview. Recall that the distribution of a function of random variables $y = g(X)$ is given by $f(g^{-1} (y)) \det{J}$, where $f$ is the original pet and $J$ is the Jacobian of the inverse transformation $g^{-1}$. Since f is constant, we need only consider the Jacobian. Since we have
$$x_i = \sin(\phi_1)\sin(\phi_2)\ldots\sin(\phi_{i-1})\cos(\phi_i)$$
Conveniently, the Jacobian is upper triangular, so we only need the product of the diagonal entries.
$$\frac{\partial x_i}{\partial \phi_i} = -\prod_{j=1}^{i} \sin(\phi_j)$$
Taking the absolute value of the product of all the entries, we get
$$f(\vec{\phi}) = \prod_{i=1}^{n} \sin(\phi_i)^{n-i+1} $$
Which is the answer we're looking for. As a sanity check, if we had a uniform distribution on the n-sphere, then the conditional distribution of the remaining coordinates conditioned on the first k should be the uniform distribution on the (n- k)-sphere, which we can easily verify.
The normalization constant is unchanged by the coordinate transformation, so all that's left is to divide by the volume of the n-sphere.
