How do you actually calculate inverse $\sin, \cos, $ etc. ? I started to wonder, how does one actually calculate the $\arcsin, \arccos, $ etc. without a calculator? For example I know that: 
$$\arccos(0.3) = 72.54239688^{\circ}$$
by an online calculator, but how would one calculate this with a pencil and paper? How is it done? :) I can't remember any math teacher talk about this? How is it actually implemented? 
Thnx for any help =) 
 A: When i was in high school a friend of mine asked our math theater this question and he told us that if he couldn't use any calculator, because they were not available so easily as today or for any other reason he was used to use Trigonometric tables
In your example you can pass through the list of angles and you see that 
$$\cos(\theta) = 0.3 \implies \theta \space \text{~} \space 72$$
He also said that obviously this method is not precise, but it gives a sufficient approximation.
P.S. The first trigonometric table was created in the second century before Christ by Hipparchus
A: Basically you can use infinite series to calculate approximation of inverse trigonometric functions.
$$
\arcsin z = z+ \left( \frac 12 \right) {z^3 \over 3} + \left( {1 \cdot 3 \over 2 \cdot 4} \right){z^5 \over 5} + \left( {1 \cdot 3 \cdot 5 \over 2 \cdot 4 \cdot 6} \right){z^7 \over 7}\ +\ ... \; = \sum_{n=0}^\infty {\binom{2n}{n}z^{2n+1} \over 4^n(2n+1)}; \;\;\; |z| \le 1
$$
$$
    \arccos z = \frac \pi2 - \arcsin z = \frac \pi2 - \sum_{n=0}^\infty {\binom{2n}{n}z^{2n+1} \over 4^n(2n+1)}; \;\;\; |z| \le 1
$$
$$
    \arctan z = z-{z^3 \over 3}+{z^5 \over 5}-{z^7 \over 7}\ +\ ... \ =  \sum_{n=0}^\infty{(-1)^nz^{2n+1} \over 2n+1}; \;\;\;|z|\le1, \; z\neq i, -i
$$
A: You don't, because you actually can't (except a few specific values). And that applies to most real functions, actually...
You can, however, calculate some good approximation, for example with appropriate polynomials.
