Finding Locally flat coordinates on a unit sphere I know this is more of a math question, but no one in the Mathematics community was able to give me an answer, and since physicists are familiar with General Relativity, I thought I might get an answer.
Imagine a unit sphere and the metric is:
$$ds^2 = d\theta ^2 + \cos^2(\theta) d\phi^2$$
I want to find Locally Flat Coordinates (I think they're called Riemann Normal Coordinates) on the point $(\frac{\pi}{4}, 0)$, so what I need are coordinates such that the metric would reduce to the Kronecker Delta and the Christoffel Symbols should vanish. I start by the following translation:
$$\theta' = \theta - \frac{\pi}{4}$$
then do the following substitution by guessing:
$$\frac{f(\theta')}{\cos(\theta)} d\phi' = d\phi$$
And the condition is $f(0)$ should be 1, so the metric becomes:
$$ds^2 = d\theta' + f^2(\theta')d\phi'$$
And it is a matter of finding $f(\theta')$. I calculate the Christoffel Symbols:
$$\Gamma^{\lambda}_{\mu\nu} = \frac{1}{2} g^{\lambda \alpha}(\partial_{\mu}g_{\alpha \nu} + \partial_{\nu}g_{\mu \alpha} - \partial_{\alpha}g_{\mu \nu})$$
And make them vanish.
So what I get is:
$$\frac{f'(0)f(0)}{f^2(0)} = 0$$
Obviously, $f(\theta')=\cos(\theta')$ is a solution which is the thing I know is correct. However, there are infinite functions that satisfy the above conditions. Are all of these functions eligible to make the new coordinates Riemann normal coordinates?
 A: starting with components  of the unit sphere :
\begin{align*}
 &\begin{bmatrix}
   x \\
   y \\
   z \\
 \end{bmatrix}=\left[ \begin {array}{c} \cos \left( \phi \right) \sin \left( \theta
 \right) \\ \sin \left( \phi \right) \sin \left(
\theta \right) \\ \cos \left( \theta \right)
\end {array} \right]
\end{align*}
from here
\begin{align*}
 &\begin{bmatrix}
   dx \\
   dy \\
   dz \\
 \end{bmatrix}=\underbrace{\left[ \begin {array}{cc} \cos \left( \phi \right) \cos \left( \theta
 \right) &-\sin \left( \phi \right) \sin \left( \theta \right)
\\ \sin \left( \phi \right) \cos \left( \theta
 \right) &\cos \left( \phi \right) \sin \left( \theta \right)
\\ -\sin \left( \theta \right) &0\end {array}
 \right]}_{\mathbf J}\, \left[ \begin {array}{c} d\theta \\ d\phi
\end {array} \right]
\end{align*}
and the metric
\begin{align*}
 &\mathbf{G}=\mathbf J^T\,\mathbf J=\left[ \begin {array}{cc} 1&0\\ 0& \left( \sin
 \left( \theta \right)  \right) ^{2}\end {array} \right]
\end{align*}
now we are looking for the transformation matrix $~\mathbf{T}~$ that transformed the metric to unit matrix
\begin{align*}
 &\mathbf{T}^T\,\mathbf{G}\,\mathbf T=\begin{bmatrix}
                                       1 & 0 \\
                                       0 & 1 \\
                                     \end{bmatrix}\quad\Rightarrow\quad
\mathbf{T}=\begin{bmatrix}
              1 & 0 \\
              0 & \frac{1}{\sin(\theta)} \\
            \end{bmatrix}
\end{align*}
hence
\begin{align*}
 &\begin{bmatrix}
   dx \\
   dy \\
   dz \\
 \end{bmatrix}\mapsto \underbrace{\mathbf{J}\,\mathbf T}_{\mathbf{T}_n}\,
 \left[ \begin {array}{c} d\theta \\ d\varphi
\end {array} \right]
\end{align*}
and the neue metric  is:
\begin{align*}
  &dx^2+dy^2+dz^2\mapsto d\theta^2+d\phi^2
\end{align*}
where $~\mathbf{T}_n~$ is a function of $~\theta~,\phi~$
\begin{align*}
 &\mathbf{T}_n(\theta=\pi/4~,\phi=0)=
 \left[ \begin {array}{cc} \frac 12\,\sqrt {2}&0\\ 0&1
\\  -\frac{1}{2}\,\sqrt {2}&0\end {array} \right]
\end{align*}
