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# how to write tan inverse

0
521
5

how to write tan inverse

Mar 29, 2015

#4
+20850
+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 :  $$\boxed{\ \alpha = \mathrm{atan} 2{( y_p, x_p )}\ }$$

.
Mar 29, 2015

#1
+95360
+5

arc tan is the same as inverse tan.

Just use atan

Mar 29, 2015
#2
+20850
+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

Mar 29, 2015
#3
+95360
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 ?

Mar 29, 2015
#4
+20850
+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
+95360
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 :)

Mar 29, 2015