Metric on sphere. In the book titled 
Space, time, and gravity by Robert Wald, I found the following, and I am quoting him,
"the metric of the sphere is
$ds^2=R^2[(d\psi)^2+cos^2\psi(d\phi)^2].$
1-where $ds$ denotes the distance between two nearby points , $d\psi$ their difference in latitude, $d\phi$ their difference in longitude, and $R$ the radius of the sphere."
2- Is the above formula correct? I think cosine should be replaced to sine.
3- also, is the exponent 2 on $ds$  square?, or just the symmetric product $dsds$? since I am confused why he squared the other two quantities i,e. $(d\theta)^2$ why not just write the metric the normal way.
4- and is there a way to explain to a laymans person what's the two here $ds^2$ without confusing them that it's a square?
5- if I give you for example that between tow points the difference in longitude and in latitude is $\pi$, and radius is 3, then the distance between the two points is $18\pi^2$ since by these information we get
$ds^2= 9[\pi^2+( cos^2(\pi) ) \pi^2]=18\pi^2$?.
thank you.

*

*"Nearby" here means "infinitesimally nearby", or "close enough that the deviation from flat space doesn't matter". Mathematically, it means an implied limit to zero.

*The formula is correct. Note that latitude is zero on the equator and maximal at the north pole, with latitude of the north pole being $\pi/2$. You are probably more familiar with the spherical coordinates commonly used in physics, where instead of the latitude $\psi$ (angle from the equator), the angle from the north pole, $\theta=\frac\pi2-\psi$, is used. For spherical coordinates, you indeed get a sine, as $\cos\psi=\cos(\frac\pi2-\theta)=\sin \theta$.

*The exponent is square (but note my remark on point 1). Basically, the formula is Pythagoras: When looking at sufficiently small distances, you cannot distinguish the sphere from an Euclidean plane (again, mathematically this informal description is formalized by the limit process), and therefore you can apply the Pythagoras rule: Distance squared ($ds^2$) equals "vertical" distance squared ($(R\,\mathrm d\psi)^2$) plus "horizontal" distance squared ($(R\cos\theta\,\mathrm d\phi)^2$).

*It's a square of an extremely small quantity. Not mathematically exact, but close enough to understand. To make it rigorous, you cannot get around using mathematics anyway.

*A difference of $\pi$ is not small by any measure. However, if the difference in both quantities is $10^{-6}$ each, then simply inserting in the formula will indeed give you a very good approximation to the actual distance.
