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This might be some kind of foundational assertion. Given any angle $\psi$ and $x\in\Bbb Z^+$ I'd like to see proof that:

$$\psi = \psi + 2x\pi$$

It's hard cause numerically this isn't true, but in the sense of angles it is. I specifically need this to make my proof based on more than my on personal assertion here: https://math.stackexchange.com/a/2144499/194115

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    What's your definition of angle? – 2017-02-14
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    You can define an "angle" as a point on the unit circle, in which case you can "calculate an angle" by plugging the numerical value into the sine and cosine functions. The sine and cosine functions both have period $2\pi$. – 2017-02-14
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    @user4894 Well I'm using it to define the orientation of an object within a plane. – 2017-02-14

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You could define an equivalence relation on the set of angles: $\alpha \sim \beta$ if $\alpha = \beta + 2\pi n$ for some integer $n$

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Mathematically, it is not true that $\psi = \psi + 2x\pi$.

What is true is that, for any trig function, since they have a period of $2\pi$ (some of them, like $\tan$, $\pi$), that $f(\psi) = f(\psi + 2x\pi)$.