The spherical polar coordinate are $x^a=(r,\theta,\phi)$ and the cylindrical polar coordinates are $x'^a=(r',\theta',z')$. I have to convert from spherical metric tensor to cylindrical metric tensor. I know the equations for converting from cylindrical to spherical coordinates are
$$ \begin{split} &r=\sqrt{r'^2+z'^2} \\ &\theta=\theta' \\ &\phi=\cos^{-1}\left(\frac{z'}{\sqrt{r'^2+z'^2}}\right), \end{split} $$
and the spherical metric tensor is
$$ g_{ab} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2\sin^2\theta \end{bmatrix}. $$
Using
$$ g_{cd}'(x') = g_{ab}(x)\frac{\partial x^a}{\partial x'^c}\frac{\partial x^b}{\partial x'^d}, $$
I can find the elements of the cylindrical metric tensor. However, when I calculate the first element of the cylindrical metric tensor I have
$$ \begin{split} g_{11}' = g_{r'r'}' &= g_{11}(\frac{\partial r}{\partial r'})^2 + g_{22}(\frac{\partial \theta}{\partial r'})^2 + g_{33}(\frac{\partial \phi}{\partial r'})^2 \\ &= 1 \cdot (\frac{\partial r}{\partial r'})^2 + r^2 \cdot (\frac{\partial \theta}{\partial r'})^2 + r^2\sin^2\theta \cdot (\frac{\partial \phi}{\partial r'})^2 \\ &= 1 \cdot (\frac{\partial r}{\partial r'})^2 + (r'^2+z'^2) \cdot (\frac{\partial \theta}{\partial r'})^2 + (r'^2+z'^2)\sin^2\theta' (\frac{\partial \phi}{\partial r'})^2 \\ &= (\frac{r'}{\sqrt{r'^2+z'^2}})^2 + (r'^2+z'^2)\sin^2\theta' (\frac{r'z'}{\sqrt{(r'^2+z'^2)^3}\sqrt{1-\frac{z'^2}{r'^2+z'2}}})^2 \\ &= \frac{1}{r'^2+z'^2}(r'^2+\sin^2\theta'z'^2). \end{split} $$
I am confused by the $\sin^2\theta'$. I know that the cylindrical metric tensor element is $g_{r'r'}'=1$, and it is easy to see that $\sin^2\theta'=1$ satisfied when $\theta=\pi/2$. But why $\sin^2\theta'=1$? How does $\sin^2\theta'$ disappear?