## From polar sytem to cartesian system

We have for point P: $OP = r ~,~~P''O P = \alpha$

Looking right triangle P”OP we can determine: $x_P=r \sin{\alpha}~, ~~ y_P = r \cos{\alpha}$.

## From cartesian system to polar system

If we consider again the right triangle P”OP we have:

$\tan{\alpha}= \dfrac{x_P}{y_P}~\rightarrow ~ \alpha = \arctan{\dfrac{x}{y}}$ $r = \dfrac{x}{\sin{\alpha}} ~=~ \dfrac{y}{\cos{\alpha}}$