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?

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*}
