Going from polar coor. to cartesian coor.

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SUMMARY

The discussion focuses on converting polar coordinates back to Cartesian coordinates, specifically addressing the challenge of determining the correct angle when both coordinates are negative. The solution involves using the atan2 function, which accounts for the quadrant of the point and provides a unique angle across the full range of -π to +π. The participants confirm that adding π to the angle derived from atan(b/a) is necessary when both a and b are negative, ensuring accurate representation in the third quadrant.

PREREQUISITES
  • Understanding of polar and Cartesian coordinate systems
  • Familiarity with trigonometric functions, specifically sine and cosine
  • Knowledge of the atan and atan2 functions
  • Basic concepts of quadrants in the Cartesian plane
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  • Research the implementation and usage of the atan2 function in various programming languages
  • Explore the mathematical properties of polar coordinates and their conversion to Cartesian coordinates
  • Study the implications of quadrant determination in trigonometric calculations
  • Learn about handling edge cases in coordinate transformations, such as points on the axes
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Mathematicians, computer scientists, and software developers working with graphical representations or simulations that require coordinate transformations.

Marioqwe
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Hello, I recently run into a problem. Let's say I have the point (a,b) and (-a,-b). The, I know that θ_1 = atan(b/a) and θ_2 = atan((-b)/(-a)) = θ_1.

But, what if I want to go back to Cartesian coordinates? If I assume r = 1,
a = cos(θ_1) and b = sin(θ_1) while
-a = cos(θ_2) and b = sin(θ_2).

I am sure this is very simple and it has to do with the fact that the range of atan is (-π/2,π/2). But is there a way of getting back the -a? Could I just add π to the angle whenever a and b are negatives?
 
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Atan is multivalued, so you need to use more information to get the angle.
 
Marioqwe said:
Hello, I recently run into a problem. Let's say I have the point (a,b) and (-a,-b). The, I know that θ_1 = atan(b/a) and θ_2 = atan((-b)/(-a)) = θ_1.

But, what if I want to go back to Cartesian coordinates? If I assume r = 1,
a = cos(θ_1) and b = sin(θ_1) while
-a = cos(θ_2) and b = sin(θ_2).

I am sure this is very simple and it has to do with the fact that the range of atan is (-π/2,π/2). But is there a way of getting back the -a? Could I just add π to the angle whenever a and b are negatives?

In a word, yes. If a and b are both negative, then the point is in the third quadrant and you would need to add ∏ to atan(b/a) to derive θ_2. [Alternately you could decide to use a non-canonical polar representation with a negative value for r].

And yes, ignoring your sign omission, -a = cos(θ_2) and -b = sin(θ_2).

Some math libraries have a two-argument "atan2" function that figures the quadrants out for you so that the range of the atan2 is the full -∏ (exclusive) to +∏ (inclusive). This function also avoids the divide by zero problem for points on the y axis.

http://en.wikipedia.org/wiki/Atan2
 
I can certainly use atan2. Thank you.
 
Matt Benesi said:
atan isn't multivalued- usually the format is: atan(y/x);

atan2 is multivalued... (wikipedia link to definition of atan2 in terms of extension of atan function)

Multivalued means there is more than one value for the given argument. Atan, like asin and acos are multivalued. Atan2 is also multivalued, but as long as you stay within an interval of 2π you will get a unique answer. For atan the interval is π, so you need to know the sign of y or x to get the right value.
 

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