How to convert back from $\cos$? Let's say I have a $\cos(1000)$ which is $0.562379076290703$.
Now what I want to do is to get number $1000$ back. How can I convert back from $0.562379076290703$?  
I tried $\arccos$ but it isn't what I need.
 A: Recall that for $x\in[0,\pi]$
$$\arccos (\cos (x))= x$$
and therefore
$$\arccos (\cos (x+2k\pi))= x$$
then we can't reconstruct the original value for $X=x+2k\pi$.
Note that in this particular case since $X=1000=2\cdot 159\cdot \pi+x$ with $x\approx 0.9735$, we have
$$X=\arccos (\cos (1000))+2\cdot 159\cdot \pi$$
A: The range of $\cos^{-1}$ is from $0$ to $\pi$. $\cos(1000 \pm 2\pi)$ will be equal to $0.562379076290703$ so there is no way to get the specific number $1000$ "back". It is "lost" once you take the $\cos$.
A: $\cos1000^\circ= \cos280^\circ = \cos 1360^\circ$ etc., 
it repeats every $360^\circ$ or $2\pi$ so $\arcsin$ will always give you the lowest possible value
A: As the others have mentioned, trig functions are periodic. In this case, you’re using cosine.
$$\cos (x-4\pi) = \cos (x-2\pi) = \cos x = \cos (x+2\pi) = \cos (x+4\pi)$$
$$\cos (x-720) = \cos (x-360) = \cos x = \cos (x+360) = \cos (x+720)$$
Recall that angles can be thought of in terms of circles. A full revolution is $360°$ or $2\pi$ radians. Once a full revolution is done, the angles start over, which is why the statements above are true.
Hence, any angle repeats an infinite number of times by repeatedly adding/subtracting $360°$ or $2\pi$ radians. (For example, $30°$ is equivalent to $390°$, $750°$, etc. It is also equivalent to $-330°$, $-690°$, etc.) This can be generalized as follows for all $n \in \mathbb{Z}$. (All integer values of $n$, positive and negative.)
$$\cos (x+2\pi n) = \cos x$$
$$\cos (x+360n) = \cos x$$
When working with functions and their inverses, it is important to remember their domains and ranges.
The domain and range of $\cos x$ are $\color{blue}{0 \leq x \leq \pi}$ and $\color{purple}{-1 \leq y \leq -1}$ respectively.
For $\arccos x$, the domain and range will be inverted (since it’s an inverse): $\color{purple}{-1 \leq x \leq -1}$ and $\color{blue}{0 \leq y \leq \pi}$
As an example, if you calculate $\cos \big(\frac{13\pi}{6}\big)$, you will get the answer $\frac{\sqrt{3}}{2}$, but if you enter $\arccos \frac{\sqrt{3}}{2}$, you will get $\frac{\pi}{6}$. This is because $\arccos x$ will return the angle within the range $0-2\pi$.
The same applies to your case. $1000$ radians is clearly out of the range of the $\arccos$ function, so it will return the value of that angle WITHIN that range.
As for how to get the $1000$ back, you just use the fact that $x = x+2\pi n$.
After entering $\arccos(\cos 1000)$, the value obtained is $0.973936$. This is the particular measure of the angle within the range.
But you know that $0.973936 = 0.973936 +2\pi n$. With the following equation, you can solve for $n$.
$$0.973936+2\pi n= 1000$$
$$n = 159$$
Therefore, your initial angle of $1000$ is really just $0.973936\cdot 159\cdot 2\pi n$.
