I want to express a point on a map in radian spherical coordinates. By Google maps, this location is north of Canton, MS, USA just a few hundred feet from US 51.
In radian spherical coordinates, $(?,-\pi/2,1) = (?, 3\tau/4,1)$. The first one is radius of Earth, the second is radians east, and the third is the angle from N. Pole.
In other words, how to describe the radius at the point which is $-\pi/2$ or $3\tau/4$ is 90 degrees west. And with the angle from North Pole as 1 or $(\pi/2-1)$rad north = $(\tau/4 -1)$ rad north which is 32.7042204869 degrees N.
I know the Earth is not a sphere, but nearly spherical. I also know that the equator had one radius and the north pole another. Which means that the radius at this location is in between these two radii. Which of the two should be used as radius = 1? With the north radian measure, being at 1, does that mean it is using radius = 1 of the pole, equator, or what is at the questioned location?
A one-line answer is: "None of the above."
A slightly more detailed answer is that the assumption that the north radian measure of this point is $1$ is inconsistent with the Earth being a spheroid with significant positive flattening.
A good reference is http://williams.best.vwh.net/ellipsoid/ellipsoid.html.
The gist of this is that if you assume a spheroidal but non-spherical Earth, with a polar radius less than its equatorial radius, the point with geodetic coordinates N$32.704220$, W$90.000000$ will be at an angle greater than $57.29578$ degrees from the north pole in spherical coordinates. The angle in spherical coordinates is $\angle POC = \pi/2 - \angle POD$ in the figure below (from http://williams.best.vwh.net/ellipsoid/node1.html), which is greater than $\pi/2-\angle P'OD = \pi/2 - \theta$, which in turn is greater than $\pi/2 - \phi$, assuming that $\phi$ is your north coordinate converted to radians.
If the equatorial radius is $a$ and the polar radius is $b$, then the angles $\phi$ and $\theta$ in the figure are related by the equation
$$\tan\theta = \frac ba \tan\phi.$$
Having used this to find $\theta$, we find the radius, $|OP|$, as follows:
$$|OP| = \sqrt{|OD|^2 + |DP|^2} = \sqrt{a^2 \cos^2\theta + b^2 \sin^2\theta}.$$
To find the correct third coordinate in spherical coordinates, namely, $\angle POC$, we use
$$\cot\angle POC = \frac ba \tan\theta = \frac{b^2}{a^2} \tan\phi.$$
If the original source of this question expected the third coordinate of the spherical triple to be $1$, then either they were expecting you to assume a spherical Earth after all, or they were misinformed about the nature of geodetic coordinates on the actual Earth.