Interpolating Between 2 Angles I'm trying to understand how this works, and mathematically I'm having difficulty.
Given 2 angles between $(-2\pi, 2\pi)$: $\theta$ and $\phi$ I want to interpolate between them by the ratio: r.
My problem is this, when interpolating between say 0 and $15\pi/8$ I don't want to get $15\pi/16$ I want to get $31\pi/16$ or $-\pi/16$.
Some math for this has been presented here: https://stackoverflow.com/a/14498790/2642059 But I can't prove to myself that for any valid $\theta$-$\phi$ combination this is workable. Can someone help me? The equation is:
$$\phi + ((((\theta-\phi)\%2\pi+3\pi)\%2\pi)-\pi)r$$
 A: From origine 0 to angle
If $0\le\theta\le\pi$ then interpolation from $0\to\theta$ is $r\theta$.
If $-\pi\le\theta\le0$ then interpolation from $0\to\theta$ is also $r\theta$ (this is your short path requirement).
If $\pi\le\theta\le2\pi$ then $-\pi\le(\theta-2\pi)\le0$ and interpolation from $0\to\theta$ is then $r(\theta-2\pi)$.
If $-2\pi\le\theta\le-\pi$ then $0\le(\theta+2\pi)\le\pi$ and interpolation from $0\to\theta$ is then $r(\theta+2\pi)$.
Between two angles
This is very similar, just shift the origin by $\theta_0$ and use $\theta=\theta_1-\theta_0$
If $\theta_0\le\theta_1\le\theta_0+\pi$ then interpolation $\theta_0+r(\theta_1-\theta_0)$.
If $\theta_0-\pi\le\theta_1\le\theta_0$ then interpolation is $\theta_0+r(\theta_1-\theta_0)$.
If $\theta_0+\pi\le\theta_1\le\theta_0+2\pi$ then interpolation is $\theta_0+r(\theta_1-\theta_0-2\pi)$.
If $\theta_0-2\pi\le\theta_1\le\theta_0-\pi$ then interpolation is $\theta_0+r(\theta_1-\theta_0+2\pi)$.
Conclusion
Now if we want to be in the cases exposed before we need to have $\theta_0\in[-\pi,\pi]$ and $\theta_1\in[-\pi,\pi]$ so that the difference $(\theta_1-\theta_0)\in[-2\pi,2\pi]$
So we do : 
$\theta_0:=(\theta_0\mod 2\pi)$ then if $(\theta_0>\pi)$ then $\theta_0:=\theta_0-2\pi$
$\theta_1:=(\theta_1\mod 2\pi)$ then if $(\theta_1>\pi)$ then $\theta_1:=\theta_1-2\pi$
$\varepsilon=sgn(\theta_1-\theta_0)$ [sign function, -1 if <0 and +1 if >0 else 0]
Our angles are now correct and we can interpolate :
If $|\theta_1-\theta_0|\le\pi$ then $\theta(r)=\theta_0+r(\theta_1-\theta_0)$
If $|\theta_1-\theta_0|\ge\pi$ then $\theta(r)=\theta_0+r(\theta_1-\theta_0-2\varepsilon\pi)$
Note that for $(\theta_1-\theta_0)=\pi$ it's up to you to chose the side, your short path requirement introduce a discontinuity there.

Now let's have a look at the formula your proposed :
$x=[(\theta_1-\theta_0) \mod 2\pi]$ and $0\le x\lt 2\pi$
We know that $a\mod b=a-b\lfloor\frac ab\rfloor$ 
In the case $0\le x\lt \pi$ then $1<\frac 32\le \frac{x+3\pi}{2\pi}<2$
$[(x+3\pi)\mod 2\pi]-\pi=(x+3\pi)-2\pi\lfloor\frac{x+3\pi}{2\pi}\rfloor-\pi=(x+3\pi)-2\pi*1-\pi=x$
And we get back our formula $\theta(r)=\theta_0+rx$
In the case $\pi\le x< 2\pi$ then $2\le \frac{x+3\pi}{2\pi}<\frac 52<3$
$[(x+3\pi)\mod 2\pi]-\pi=(x+3\pi)-2\pi\lfloor\frac{x+3\pi}{2\pi}\rfloor-\pi=(x+3\pi)-2\pi*2-\pi=(x-2\pi)$
And we get back our formula $\theta(r)=\theta_0+r(x-2\pi)$
This is quite a clever trick :-)
A: You want an interpolation the shortest distance between $\theta$ and $\phi$, which means that we don't care about r.
What you're really asking is for a proof that: $-\pi\leq(((\theta-\phi)\%2\pi+3\pi)\%2\pi)-\pi\leq\pi$
We can simplify that to: $0\leq((\theta-\phi)\%2\pi+3\pi)\%2\pi\leq2\pi$
This is guaranteed to be less than $2\pi$ because of the modulo operator, so we now just need to know that: $0\leq(\theta-\phi)\%2\pi+3\pi$
Which can again be simplified to: $-3\pi\leq(\theta-\phi)\%2\pi$
This is guaranteed to be true independent of the value of $\theta-\phi$ because: $\forall x:-2\pi<(x)\%2\pi$
This proves that the angle will always be the shortest angle between $\theta$ and $\phi$.
It also needs to be true that this is the angle such that: $(((\theta-\phi)\%2\pi+3\pi)\%2\pi)-\pi=\theta-\phi$
Which simplifies to: $((\theta-\phi)\%2\pi+3\pi)\%2\pi=\theta-\phi+\pi$
I'm going to make the basic assertion that: $\forall x\in\Bbb Z^+:\psi=\psi+2x\pi$ And by extension taking the $2\pi$ modulo of any angle will simply result in the same angle.
Therefore: $(\theta-\phi)\%2\pi+3\pi=\theta-\phi+\pi$
Which simplifies to: $(\theta-\phi)\%2\pi=\theta-\phi-2\pi$
By applying my assertion to both sides we get: $\theta-\phi=\theta-\phi$
