Representation of trigonometric functions in terms of a cross product Let us say that I define a vector in 3D given by
$$ \vec{u}=(\sin\theta \cos\phi, \sin\theta\sin\phi,\cos\theta)$$
I can write $\sin\theta$ in terms of a cross product of the derivatives of the vector (verified by hand and with Mathematica):
$$\vec{u}\cdot \frac{\partial \vec{u}}{\partial \theta}\times \frac{\partial \vec{u}}{\partial \phi} =\sin\theta$$
Is there a similar way to write other trigonometric functions; i.e., $\cos\theta$ or $\tan\theta$? In other words, how can I write these trigonometric functions in terms of a cross product of derivatives of $\vec{u}$ with respect to $\theta$ and $\phi$? Can I do the same for $\sin\phi$, $\cos\phi$, or $\tan\phi$?
 A: $\vec{u}$ is a unit vector consisting of a rotation of $\phi$ around the $z$ axis, with $\phi=0$ corresponding to the $y$ axis, resulting in a unit vector $u'$ in the $xy$ plane, followed by a rotation $\theta$ of $u'$ about the axis perpendicular to the $z-u'$ plane resulting in the unit vector $u$.
$$
\vec{a}\equiv
\frac{\partial \vec{u}}{\partial \phi}=
\begin{pmatrix}
-\sin\theta\sin\phi
&
\sin\theta\cos\phi
&
0
\end{pmatrix}
$$ is a vector of length $\Vert \vec{a}\Vert=\sin\theta$ in the $xy$ plane perpendicular to $u$
$$
\vec{u}\cdot\frac{\partial \vec{u}}{\partial \phi}=0
$$
and $$
\vec{b}\equiv
\frac{\partial \vec{u}}{\partial \theta}
=
\begin{pmatrix}
\cos\theta\cos\phi
&
\cos\theta\sin\phi
&
-\sin\theta
\end{pmatrix}
$$ is a vector perpendicular to $\vec{a}$ of length $\Vert \vec{b}\Vert=1$. When you take the cross product of these vectors, you get a vector
$$
\vec{c}=\vec{a}\times\vec{b}=\frac{\partial \vec{u}}{\partial \theta}\times\frac{\partial \vec{u}}{\partial \phi}\\
=
\begin{pmatrix}
\sin^2\theta\cos\phi
&
\sin^2\theta\sin\phi
&
\sin\theta\cos\theta
\end{pmatrix}
\equiv
\sin\theta \vec{u}
$$
which is perpendicular to both of the derivatives, hence must be in the same direction as $\vec{u}$, and has length $\Vert\vec{c}\Vert=\sin\theta$. When you take the dot product of that vector with $u$, because $\vec{b}$ and $\vec{u}$ are unit vectors, and $\vec{a}$ has length $\sin\theta$, the result comes out to $\sin\theta$.
So to answer your question, of course you have a formula for $\sin\theta$ so you can get $\cos\theta=\sqrt{1-\sin^2\theta}$ and so on, but you can't get $\phi$. Vectors $\vec{a}$ and $\vec{b}$ and $\vec{u}$ are all orthogonal, and  the only angle which occurs is the angle from the length of $\vec{a}$, $\theta$.
