Originally Posted by

**Ackbeet** You're correct. However, that's not quite what they're doing. For one thing, they're dividing by the length. So, if you first compute

$\displaystyle \displaystyle\frac{d\mathbf{r}}{d\theta}=\begin{bm atrix}-r\sin(\theta)\sin(\phi)\\ r\cos(\theta)\sin(\phi)\\ 0\end{bmatrix},$ and then find its length, which is

$\displaystyle \displaystyle\left|\frac{d\mathbf{r}}{d\theta}\rig ht|=r\sqrt{\sin^{2}(\theta)\sin^{2}(\phi)+\cos^{2} (\theta)\sin^{2}(\phi)}=r|\sin(\phi)|=r\sin(\phi),$

on account of the range of $\displaystyle \phi$ (the polar angle on this website), then you'll find that

$\displaystyle \hat{\mathbf{\theta}}=\dfrac{\dfrac{d\mathbf{r}}{d \theta}}{\left|\dfrac{d\mathbf{r}}{d\theta}\right| }=\dfrac{1}{r\sin(\phi)}\begin{bmatrix}-r\sin(\theta)\sin(\phi)\\ r\cos(\theta)\sin(\phi)\\ 0\end{bmatrix}=\begin{bmatrix}-\sin(\theta)\\ \cos(\theta)\\ 0\end{bmatrix},$

as they have. Does that make it clearer, perhaps?