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how to write tan inverse

Guest Mar 29, 2015

Best Answer 

 #4
avatar+20040 
+5

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 :  $$\boxed{\ \alpha = \mathrm{atan} 2{( y_p, x_p )}\ }$$

heureka  Mar 29, 2015
 #1
avatar+93691 
+5

arc tan is the same as inverse tan.

Just use atan

Melody  Mar 29, 2015
 #2
avatar+20040 
+5

 

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

heureka  Mar 29, 2015
 #3
avatar+93691 
0

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 ?

Melody  Mar 29, 2015
 #4
avatar+20040 
+5
Best Answer

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 :  $$\boxed{\ \alpha = \mathrm{atan} 2{( y_p, x_p )}\ }$$

heureka  Mar 29, 2015
 #5
avatar+93691 
0

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 :)

Melody  Mar 29, 2015

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