Direction Angle of 2D Vector in Terms of Arctangent

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Theorem

Let $\mathbf a$ be a vector quantity embedded in a Cartesian plane $P$ expressed in component form as:

$\mathbf a = x \mathbf i + y \mathbf j$


Let $\theta$ denote the direction of $\mathbf a$.

Then:

$\theta = \begin{cases}

\map \arctan {\dfrac y x} & : x > 0 \\ \map \arctan {\dfrac y x} + \pi & : x < 0 \text{ and } y \ge 0 \\ \map \arctan {\dfrac y x} - \pi & : x < 0 \text{ and } y < 0 \\ \dfrac \pi 2 & : x = 0 \text{ and } y > 0 \\ -\dfrac \pi 2 & : x = 0 \text{ and } y < 0 \\ \text{undefined} & : x = 0 \text { and } y = 0 \end{cases}$

where:

$\arctan$ denotes the real arctangent function, defined on the real interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$
$\theta$ is conventionally measured from the positive direction of the $x$-axis in the interval $\hointl {-\pi} \pi$.


Proof

Let $\mathbf a$ be such that one of the following holds:

$\mathbf a$ is in Quadrant $\text{I}$ or Quadrant $\text{IV}$
$\mathbf a$ is on the positive direction of the $x$-axis.

Then:

$x > 0$

and:

$-\dfrac \pi 2 < \theta < \dfrac \pi 2$

The components of $\mathbf a$ form the legs of a right triangle where:

\(\ds \tan \theta\) \(=\) \(\ds \frac y x\) Definition of Tangent Function
\(\ds \leadsto \ \ \) \(\ds \map \arctan {\tan \theta}\) \(=\) \(\ds \map \arctan {\frac y x}\) can take arctangent of both sides as $-\dfrac \pi 2 < \theta < \dfrac \pi 2$
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \theta\) \(=\) \(\ds \map \arctan {\frac y x}\) Composite of Bijection with Inverse is Identity Mapping

$\Box$


Let $\mathbf a$ be such that one of the following holds:

$\mathbf a$ is in Quadrant $\text{II}$
$\mathbf a$ is on the negative direction of the $x$-axis.

Then:

$x < 0$
$y \ge 0$

and:

$\dfrac \pi 2 < \theta \le \pi$
\(\ds \tan \theta\) \(=\) \(\ds \frac y x\) Definition of Tangent Function
\(\ds \leadsto \ \ \) \(\ds \tan \paren {\theta - \pi}\) \(=\) \(\ds \frac y x\) Tangent Function is Periodic on Reals
\(\ds \leadsto \ \ \) \(\ds \map \arctan {\tan \paren {\theta - \pi} }\) \(=\) \(\ds \map \arctan {\frac y x}\) $\dfrac \pi 2 < \theta \le \pi \implies -\dfrac \pi 2 < \theta - \pi \le 0$, can take arctangent of both sides as $-\dfrac \pi 2 < \theta - \pi < \dfrac \pi 2$
\(\ds \leadsto \ \ \) \(\ds \theta - \pi\) \(=\) \(\ds \map \arctan {\frac y x}\) Composite of Bijection with Inverse is Identity Mapping
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \theta\) \(=\) \(\ds \map \arctan {\frac y x} + \pi\) Add $\pi$ on both sides

$\Box$


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Let $\mathbf a$ be in Quadrant $\text{III}$.

Then:

$x < 0$
$y < 0$

and:

$-\pi < \theta < \dfrac {-\pi} 2$
\(\ds \tan \theta\) \(=\) \(\ds \frac y x\) Definition of Tangent Function
\(\ds \leadsto \ \ \) \(\ds \tan \paren {\theta + \pi}\) \(=\) \(\ds \frac y x\) Tangent Function is Periodic on Reals
\(\ds \leadsto \ \ \) \(\ds \map \arctan {\tan \paren {\theta + \pi} }\) \(=\) \(\ds \map \arctan {\frac y x}\) $-\pi < \theta < \dfrac {-\pi} 2 \implies 0 < \theta + \pi < \dfrac \pi 2$, can take arctangent of both sides as $-\dfrac \pi 2 < \theta + \pi < \dfrac \pi 2$
\(\ds \leadsto \ \ \) \(\ds \theta + \pi\) \(=\) \(\ds \map \arctan {\frac y x}\) Composite of Bijection with Inverse is Identity Mapping
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \theta\) \(=\) \(\ds \map \arctan {\frac y x} - \pi\) Substract $\pi$ on both sides

$\Box$


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Let $\mathbf a$ be on the positive direction of the $y$-axis.

Then:

$x = 0$
$y > 0$

Then the arctangent of $\theta$ is undefined.

We have:

\(\text {(4)}: \quad\) \(\ds \theta\) \(=\) \(\ds \frac \pi 2\)

$\Box$


Let $\mathbf a$ be on the negative direction of the $y$-axis.

Then:

$x = 0$
$y < 0$

Then the arctangent of $\theta$ is undefined.

We have:

\(\text {(5)}: \quad\) \(\ds \theta\) \(=\) \(\ds -\frac \pi 2\)

$\Box$


Finally, let $\mathbf a$ be the zero vector.

We have:

$x = 0$
$y = 0$

and hence:

\(\text {(6)}: \quad\) \(\ds \theta\) \(=\) \(\ds \text {undefined}\) Zero Vector has no Direction


Hence, $\theta$ can be described using the following piecewise function:

\(\ds \theta\) \(=\) \(\ds \begin{cases}

\map \arctan {\dfrac y x} & : x > 0 \\ \map \arctan {\dfrac y x} + \pi & : x < 0 \text{ and } y \ge 0 \\ \map \arctan {\dfrac y x} - \pi & : x < 0 \text{ and } y < 0 \\ \dfrac \pi 2 & : x = 0 \text{ and } y > 0 \\ -\dfrac \pi 2 & : x = 0 \text{ and } y < 0 \\ \text{undefined} & : x = 0 \text { and } y = 0 \end{cases}\)

from $(1)$, $(2)$, $(3)$, $(4)$, $(5)$, and $(6)$ and the cases in which they apply

Hence the result.

$\blacksquare$


Also defined as

Some sources define the codomain as the interval $\hointr 0 {2 \pi}$.


Also known as

This piecewise function is often presented in computer languages as $\map {\text {atan2} } {y, x}$.

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