I was reading this post Scalar product of random unit vectors and I cannot understand how the density is obtained. We start by stating that $X$ and $X'$ are uniformly distributed on the sphere — which is equivalent to saying $\cos(\phi) \sim \mathcal{U}(-1,1)$ and $\gamma \sim \mathcal{U}(0, 2\pi)$ — but then user @IosifPinelis says that $X$ has the same distribution as a standard normal vector normalized to unit length. Shouldn't it be uniform on the sphere? Also, the user sets $X' = (1, 0, \ldots, 0)$.
After that, the probability density function (PDF) is given by $$ f_R(r) = \frac{\Gamma\left(\frac{d}{2}\right)}{\sqrt{\pi}\, \Gamma\left(\frac{d-1}{2}\right)} \left(1 - r^2\right)^{\frac{d-3}{2}} \mathbf{1}_{\{|r|<1\}}, $$ where $r = X \cdot X'$.
Then, @GinPat states that the PDF of the angle $\alpha = \arccos(X \cdot X')$ can be written as $$ f(\alpha) = C(d) \sin^{d-2}(\alpha) \, \mathbf{1}_{\{0 < \alpha < \pi\}}, $$ for some normalization constant $C(d)$ given by the change of variables.
I do not understand how these expressions are derived, and particularly how setting $X' = (1, 0, \ldots, 0)$ does not cause any loss of generality in the result.
Could anyone help me clarify these points? If this question is not appropriate here, I may consider posting it on Math Stack Exchange.
