how to write tan inverse

Hallo Melody,

my english is not so good, but i start.

We have a Point P, the coordinate is ( x,y ). So we have P(x,y).

The question is, what is the angle from the x-axis to that Point (we call the angle also polar angle ). See: https://en.wikipedia.org/wiki/Polar_coordinate_system

The formula for the angular coordinate is :  $$\alpha = \arctan{( \frac{y_p}{x_p} )}\\
\text{ or } \alpha = \mathrm{atan( \frac{y_p}{x_p} ) }$$

But this formula does not calculate the angle correctly. We have the same angle in  the Quadrant ( I and III ) and in the Quadrant ( II and IV ).

Why?

Because $$\frac {+y_p}{+x_p} = \frac{-y_p}{-x_p} = +\frac {y_p}{x_p}$$ and $$\frac {+y_p}{-x_p} = \frac{-y_p}{x_p} = - \frac {y_p}{x_p}$$

If we divide y by x, the information about the quadrant has disappeared.

But we can see:

$$\\\small{\text{Point in the I. Quadrant $y_p > 0 $ and $ x_p > 0 $}}\\
\small{\text{Point in the II. Quadrant $y_p > 0 $ and $ x_p < 0 $}}\\
\small{\text{Point in the III. Quadrant $y_p < 0 $ and $ x_p < 0 $}}\\
\small{\text{Point in the IV. Quadrant $y_p < 0 $ and $ x_p > 0 $}}\\$$

We must correct the angular coordinate afterwards.

and if y or x is zero, we must put constants:

$$\\\small{\text{$ y_p = 0 $ and $ x_p > 0 \qquad \alpha = 0 $ }}\\
\small{\text{$ y_p > 0 $ and $ x_p = 0 \qquad \alpha = \frac{\pi}{2}$ }}\\
\small{\text{$ y_p = 0 $ and $ x_p < 0 \qquad \alpha = \pi$ }}\\
\small{\text{$ y_p < 0 $ and $ x_p = 0 \qquad \alpha = \frac{3}{2}\pi$ }}\\$$

We have succeed, there is a function which takes this work from us!

The funktion is atan2

and needs two parametres $$\small{\text{ $y_p$ and $x_p$}}$$

The new formula for the angular coordinate is :  $$\boxed{\ \alpha = \mathrm{atan} 2{( y_p, x_p )}\ }$$

arc tan is the same as inverse tan.

Just use atan

You can also use atan2, see examples below, to get the angle in the quadrant  (I, II, III, and IV):

$$\\\text{Formula:}\\
\alpha = \mathrm{atan2}\ {(\Delta y, \Delta x)}$$

Examples:

http://web2.0rechner.de/#atan2(1,1)  $$\alpha = 45\ \mathrm{degrees} \qquad \text{Quadrant I}$$

http://web2.0rechner.de/#atan2(1,-1) $$\alpha = 135\ \mathrm{degrees} \qquad \text{Quadrant II}$$

http://web2.0calc.com/#atan2(-1,-1)    $$\alpha = -135\ \mathrm{degrees} \qquad \text{Quadrant III}$$

http://web2.0rechner.de/#atan2(-1,1)  $$\alpha = -45\ \mathrm{degrees} \qquad \text{Quadrant IV}$$

Click the "=" Button in the link

Thanks Heureka, 

I have never seen this before.  I am going to try it too :))

I am trying to use acos(0.5)  your way and get the different quadrant answers but it is not working for me.

Can you show me how to do this please Heureka ?

Thanks Heureka,  I shall study this tomorrow when i am fresh. 

I though you were just giving us a new way to enter it into the web2 calc :)