Question

Is it possible to represent arctan(x,y) in terms of arctan(y/x)?

Answer 1

(x,y)?

Answer 2

arctan(x,y) is the full arc tangent function, so it gives the correct angle when you're outside the first quadrant.

Answer 3

example please sir

Answer 4

I have no idea how to answer the question, but I am curious none the less.

Answer 5

\[\arctan(\frac{-1}{-1})=\arctan(\frac{1}{1}) = \frac{\pi}{4} \\ \arctan(-1,-1)=\frac{5\pi}{4}\]

Answer 6

Why do we mess with the other one?

Answer 7

I came across this when someone made a passing remark that \[\Large r=\sqrt{x^2+y^2} \\ \Large \theta = \arctan(\frac{y}{x})\] wasn't technically correct for the transformation from cartesian to polar and was kind of shocked lol.

Answer 8

yes

Answer 9

So the arctangent we're all familiar with has range \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\), while the full version has range \([0,2\pi)\), right?

Answer 10

Yeah I believe so.

Answer 11

arctan(x,y) makes no sense...I m8 be wrong

Answer 12

A piecewise definition would work but I'm not sure what kind of cases to implement... something like
\[\arctan(x,y)=\begin{cases}\arctan\dfrac{y}{x}&\text{for }x>0\\\\
\arctan\dfrac{y}{x}+\pi&\text{for }x<0\end{cases}\]

Answer 13

when we are saying arctan(x,y).....we are pointing to a particular point in the graph

Answer 14

so it will always be same as arctan(y/x)

Answer 15

xcept for x<0

Answer 16

@SithsAndGiggles already told it I guess

Answer 17

I'm imagining something similar @SithsAndGiggles, depending on if x and y are greater than or less than 0 you add different amounts of pi. lol

Answer 18

but here we also have to check for y>0 and y<0

Answer 19

2 variables

Answer 20

so 4 equations x> y> x<y> y<x> y<x< 0

Answer 21

Hmmm that's unfortunate that it has to look so ugly. I kind of wish there was a better way.

Answer 22

and also whether xy>0 or xy<0

Answer 23

I guess it makes sense now

Answer 24

ya.so We can I guess

Answer 25

I think we can achieve symmetry if we restrict the definition to make it so that the range of \(\arctan(x,y)\) is \((-\pi,\pi]\) rather than \([0,2\pi)\).

Answer 26

Wait so WHAT IS THE PROBLEM IN THE PROBLEM???

Answer 27

I think I answered it rt??

Answer 28

Haha yeah I think that will work best to have it from -pi to pi. I am trying to find a version that's not piecewise by using absolute value and that kind of thing.

Answer 29

This is surprisingly confusing lol

Answer 30

lol yeah.......

Answer 31

no I think u shud check for >0 and<0

Answer 32

If we want some semblance of symmetry we could get it with three cases.

Something like
\[\arctan(x,y)=\begin{cases}
\arctan\dfrac{y}{x}-\pi&\text{for }(x<0)\land(y<0)\\\\
\arctan\dfrac{y}{x}&\text{for }x>0\\\\
\arctan\dfrac{y}{x}+\pi&\text{for }(x<0)\land (y>0)
\end{cases}\]