Calculation of Christoffel symbol for unit sphere We use the following parameterisation for the unit sphere: $\sigma(\theta,\phi)=(\cos\theta\cos\phi,\cos\theta\sin\phi,\sin\theta)$.
I have calculated the Christoffel symbols to be 
$\Gamma^1_{11}=\Gamma^2_{11}=\Gamma^1_{12}=0, \Gamma^1_{22}=\sin\theta\cos\theta,\Gamma^2_{22}=0$, which match the answers I am given in my notes. But when I calculate $\Gamma^2_{12}$ I get $-\sin\theta\cos\theta$, which apparently is incorrect and should be $-\tan\theta$. My reasoning was that $\Gamma^2_{12}=\sigma_\phi \cdot \sigma_{\theta\phi}=(-\cos\theta\sin\phi,\cos\theta\cos\phi,0)\cdot(\sin\theta\sin\phi,-\sin\theta\cos\phi,0)=-\sin\theta\cos\theta$. I am not sure what I am doing wrong - the same method worked for the other five symbols and I have no idea where a $\tan\theta$ term would come from. Any help would be appreciated.
 A: $\require{cancel}$
You can directly calculate the metric coefficients for this parameterization as ($x^1 = \cos\theta, x^2 = \phi$)
$$
(g_{\mu\nu}) = \pmatrix{1 & 0 \\ 0 & \cos^2\theta} ~~~\mbox{and}~~
(g^{\mu\nu}) = \pmatrix{1 & 0 \\ 0 & 1/\cos^2\theta}
$$
From this is pretty straightforward to calculate $\Gamma^{\lambda}_{\mu\nu}$
$$
\Gamma^{\lambda}_{\mu\nu} = \frac{1}{2}g^{\lambda\alpha}\left(\frac{\partial g_{\mu\alpha}}{\partial x^{\nu}}+ \frac{\partial g_{\alpha\nu}}{\partial x^{\mu}}   - \frac{\partial g_{\mu\nu}}{\partial x^{\alpha}}\right)
$$
Take $\lambda = 2$, $\mu = 1$ and $\nu = 2$
\begin{eqnarray}
\Gamma^{2}_{12} &=& \frac{1}{2}g^{2\alpha}\left(\frac{\partial g_{1\alpha}}{\partial x^{2}}+ \frac{\partial g_{\alpha2}}{\partial x^{1}} - \frac{\partial g_{12}}{\partial x^{\alpha}}\right) = \frac{1}{2}g^{22}\left(\cancelto{0}{\frac{\partial g_{12}}{\partial x^{2}}} + \frac{\partial g_{22}}{\partial x^{1}} - \cancelto{0}{\frac{\partial g_{12}}{\partial x^{2}}}\right) \\
&=& \frac{1}{2}\left(\frac{1}{\cos^2\theta}\right) \frac{\partial \cos^2\theta}{\partial  \theta} = -\tan\theta
\end{eqnarray}
You can calculate the other components the same way
$$
\Gamma_{11}^1 = \Gamma_{11}^2 = \Gamma_{12}^1 = \Gamma_{22}^2 = 0
$$
and 
$$
\Gamma_{22}^1 = \sin\theta\cos\theta
$$
A: There's another way, and I think this is what you were going for in your post, but is is completely equivalent to my other answer, the idea is to calculate
$$
\Gamma^{\lambda}_{\mu\nu} = \frac{\partial x^\lambda}{\partial \sigma^\alpha}\frac{\partial^2 \sigma^\alpha}{\partial x^\mu \partial x^\nu} \tag{1}
$$
You can obtain the inverse mapping fairly easily 
$$
\theta = \arcsin \sigma^2 \equiv x^1 ~~~\mbox{and}~~~ \phi = \arctan \left(\frac{\sigma^2}{\sigma^1}\right) \equiv x^2 \tag{2}
$$
With this we have 
\begin{eqnarray}
\Gamma^{2}_{12} &=& \frac{\partial x^2}{\partial \sigma^\alpha} \frac{\partial^2 \sigma^\alpha}{\partial x^1 \partial x^2} \\
&=& \frac{\partial x^2}{\partial \sigma^1} \frac{\partial^2 \sigma^1}{\partial x^1 \partial x^2} + \frac{\partial x^2}{\partial \sigma^2} \frac{\partial^2 \sigma^2}{\partial x^1 \partial x^2} + \frac{\partial x^2}{\partial \sigma^3} \frac{\partial^3 \sigma^1}{\partial x^1 \partial x^2} \\
&=& -\frac{\sigma^2 \sin\theta\sin\phi}{(\sigma^1)^2 + (\sigma^2)^2}-\frac{\sigma^1\sin\theta\cos\phi}{(\sigma^1)^2 + (\sigma^2)^2} \\
&=& - \tan\theta \tag{3}
\end{eqnarray}
