If I square a measure in radians, is the result measured in steradians? Originally, SI units included radians and steradians as base units (now they're considered derived). The fact that they bothered to include radians and steradians is confusing to me. I usually think of the SI system as trying to be quite minimal. For example, there's no unit for square meters; you just square the meter.
So does the fact that they listed steradians separately from radians imply that a steradian is inherently different from a squared radian?
Like... suppose you decided to measure square angles in terms of the surface area on a sphere swept by an arc of $x$ radians. Initially the area swept by $x$ would be close to $\pi x^2$, but as you approached $x \rightarrow 2\pi$ the swept area would not be increasing very fast anymore and instead of sweeping $\pi x^2 = \pi^3$ you end up sweeping $\pi 4$. Clearly this particular measure of angle area fails to simply be the square of an angle, since it ends up only increasing linearly due to wrapping around the sphere.
Another example. Suppose you defined an angle measure as "area divided by radius". This is a bad measure, since doubling the size of your sphere will double your measure. But it also wouldn't be radians squared.
Are steradians just area over radius squared, or is there something more? When can I confidently multiply measures in radians and say they are in steradians? Are they like meters, where multiplying any two values in meters gives a value in square meters (e.g. a perimeter times a diameter gives you a value measured in square meters, though it may not be the area of the shape)?
If I have an equation that multiplies two values measured in radians, is the result measured in steradians?
 A: The radian measures an angle in the plane (2 dimensional space).  The steradian measures a vertex in 3 dimensional space.  If you had a pyramid, for example, you could use radians to measure each of the angles at the vertex, and steradians to measure the vertex as a whole.
To measure a vertex in steradians, you would imagine a unit sphere with the vertex at the center, and the measure the area of the sphere inside the vertex.
There is an alternative way to measure it.  If the intersection of the vertex and the sphere forms a polygon.  If you measure the angels and sum them, and compare that to what that number would be for a polygon with the same number of sides on a plane, the difference is the measure of the vertex.
If you had a regular tetrahedron. The measure of each angle at the vertex is $\frac \pi3$ radians, but the vertex is $3\cos^{-1} \frac 13 - \pi$ steradians.
A: This works for a solution (from wikipedia):
The solid angle of a cone with apex angle $2\theta$, is the area of a spherical cap on a unit sphere
$$\Omega =2\pi \left(1-\cos {\theta }\right).$$
For small $\theta$ such that $\sin \theta \approx \theta$, this reduces to the area of a circle $2\pi\theta$.
The above is found by computing the following double integral using the unit surface element in spherical coordinates:
\begin{align}
\int _{0}^{2\pi }\int _{0}^{\theta }\sin \theta'\,d\theta'\,d\phi
&=2\pi \int_{0}^{\theta }\sin \theta '\,d\theta' \\
&=2\pi \left[-\cos \theta '\right]_{0}^{\theta } \\
&=2\pi \left(1-\cos \theta \right) \int _{0}^{2\pi }\int _{0}^{\theta}\sin \theta'\,d\theta '\,d\phi \\
&=2\pi \int _{0}^{\theta }\sin \theta'\,d\theta' \\
&=2\pi \left[-\cos \theta'\right]_{0}^{\theta} \\
&=2\pi \left(1-\cos \theta \right)
\end{align}
This means if you have an angle $x$, the solid angle created by sweeping out $x$ radians in two dimensions is $2 \pi (1-\cos(x/2))$, not $x^2$.
I was wondering the same thing because cameras give pixel IFOV in terms of radians, but the view of the pixel is actually a solid angle where the IFOV is swept out left to right and up to down. 
A: I agree that Ω = 2π[1−cos(θ)] is correct for the solid angle subtended by a cone having half-angle θ.  The angle θ can range from 0 (a point having zero solid angle) to π/2 (a hemisphere with Ω = 2π steradians) to π (the full sphere with Ω = 4π sr).  In many imaging applications, θ is small -- perhaps π/10 (a 36 degree FOV) or less.  Expanding the cos(θ) in a power series = 1 -θ2/2 + .... and keeping just the first term with θ gives Ω ≈ πθ2, which is the "angular area" subtended by the cone (compare this to Area = πr2).  So, for sufficiently small angles, the product of two orthogonal angles in radians gives a solid angle in steradians.  
A: 1 steradian is defined as the solid angle that subtends a circular spherical cap of area = r^2.  If we accept that a radian can also represent the linear measure = r, then, 1 rad^2 = 1 r^2 = 1 sr.  Likewise, the surface area of a sphere = 4 pi r^2 = 4 pi rad^2 = 4 pi sr.  See also: https://www.physlink.com/Education/Askexperts/ae174.cfm and https://www.youtube.com/watch?v=def8n4HgphU&spfreload=10
The concept of rad^2 (or deg^2) can be difficult to picture.  It doesn't make sense to simply square arc lengths in degrees or radians because we're not working on a flat surface.  A spherical square with sides of length = s rad will not have area = s^2 rad^2 (because it is not flat), but it will be close to that for small values of s (because for small s it will be close to being flat).  
