Prove that the map $F:(-\pi,\pi)\times(-\pi/2,\pi/2)\rar S^2$,
$$
F(\vp,\theta)\colon=(\cos\theta\cos\vp,\cos\theta\sin\vp,\sin\theta)
$$
is not conformal - the set $(-\pi,\pi)\times(-\pi/2,\pi/2)$ carries the canonical Euclidean metric. Find a Riemannian metric $g$ on $M\colon=(-\pi,\pi)\times(-\pi/2,\pi/2)$ such that $F$ is a local isometry. Remark: $F$ maps the line $[\theta=0]$ onto the 'equator' of $S^2$, i.e. the equator has latitude $\theta=0$.
The Jacobi matrix $DF(\vp,\theta)$ of $F$ is given by
$$
\left(\begin{array}{cc}
-\cos\theta\sin\vp&-\sin\theta\cos\vp\\
\cos\theta\cos\vp&-\sin\theta\sin\vp\\
0&\cos\theta
\end{array}\right)~.
$$
i.e.
\begin{eqnarray*}
TF(E^\vp)&=&-\cos\theta\sin\vp\,E^x+\cos\theta\cos\vp\,E^y\\
TF(E^\theta)&=&-\sin\theta\cos\vp\,E^x-\sin\theta\sin\vp\,E^y+\cos\theta\,E^z
\end{eqnarray*}
The Gramian of these tangent vectors is
$$
\left(\begin{array}{cc}
\cos^2\theta&0\\
0&1
\end{array}\right)
$$
which is in general not a multiple of the identity. Endowing $M$ with the pull back metric $g$, i.e.
$$
g(E^\vp,E^\vp)\colon=\cos^2\theta,
g(E^\theta,E^\theta)\colon=1,
g(E^\vp,E^\theta)\colon=0
$$
turns $F:(M,g)\rar S^2$ into a local isometry.