# 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 $$\tan\phi$$?

$$\vec{u}$$ 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 $$\vec{a}$$, $$\theta$$.