+0

how to write tan inverse

0
328
5

how to write tan inverse

Guest Mar 29, 2015

#4
+19207
+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
Sort:

#1
+92221
+5

arc tan is the same as inverse tan.

Just use atan

Melody  Mar 29, 2015
#2
+19207
+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
+92221
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
+19207
+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
#5
+92221
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

31 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details