Let $x(\theta,\phi)=(\cos(\theta)\sin(\phi),\sin(\theta)\sin(\phi),\cos(\phi))$ be the standard local coordinates for $\mathbb{S}^{2} \setminus \lbrace (0,0,\pm 1) \rbrace$. Let $\phi_{0} \in [0,2\pi[$.
We define :
$$E_{1}(\theta) = \frac{\partial x/\partial \theta (\theta,\phi_{0})}{\Vert \partial x / \partial \theta (\theta,\phi_{0}) \Vert} = (-\sin(\theta),\cos(\theta),0)$$
and
$$E_{2}(\theta) = \frac{\partial x/\partial \phi (\theta,\phi_{0})}{\Vert \partial x / \partial \phi (\theta,\phi_{0}) \Vert} = (\cos(\theta)\cos(\phi_{0}),\sin(\theta)\cos(\phi_{0}),-\sin(\phi_{0}))$$
$(E_{1}(\theta),E_{2}(\theta))$ is an orthonormal basis of $T_{x(\theta_{0},\phi_{0})} \mathbb{S}^{2}$.
I don't understand why
$$ E_{1}' = -\cos(\phi_{0})E_{2} - \sin(\phi_{0})E_{3} $$
where $E_{3}(\theta) = x(\theta,\phi_{0})$.
Thanks for your help.
I think it is obvious. In fact:
$$-\cos(\phi_0)E_2=-\cos(\phi_0)\left(\cos(\theta)\cos(\phi_{0}),\sin(\theta)\cos(\phi_{0}),-\sin(\phi_{0})\right)=\left(-\cos(\theta)\cos^2(\phi_0),-\sin(\theta)\cos^2(\phi_0),-1/2\sin(2\phi_0)\right)$$ and
$$-\sin(\phi_0)E_3=-\cos(\phi_0)\left(\cos(\theta)\sin(\phi_{0}),\sin(\theta)\sin(\phi_{0}),-\cos(\phi_{0})\right)=\left(-\cos(\theta)\sin^2(\phi_0),-\sin(\theta)\sin^2(\phi_0),+1/2\sin(2\phi_0)\right)$$ So:
$$-\cos(\phi_0)E_2-\sin(\phi_0)E_3=(-\cos(\theta)\times1,-\sin(\theta)\times1,0)=\left(-\sin(\theta),\cos(\theta),0\right)'=E_1'$$