Tangent of Angle in Cartesian Plane
Theorem
Let $P = \tuple {x, y}$ be a point in the cartesian plane whose origin is at $O$.
Let $\theta$ be the angle between the $x$-axis and the line $OP$.
Let $r$ be the length of $OP$.
Then:
- $\tan \theta = \dfrac y x$
where $\tan$ denotes the tangent of $\theta$.
Proof
Let a unit circle $C$ be drawn with its center at the origin $O$.
Let a tangent line be drawn to $C$ parallel to $PS$ meeting $C$ at $R$.
Let $Q$ be the point on $OP$ which intersects this tangent line.
$\angle OSP = \angle ORQ$, as both are right angles.
Both $\triangle OSP$ and $\triangle ORQ$ share angle $\theta$.
By Triangles with Two Equal Angles are Similar it follows that $\triangle OSP$ and $\triangle ORQ$ are similar.
Thus:
Then:
\(\ds \frac y x\) | \(=\) | \(\ds \frac {SP} {OS}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {RQ} {OR}\) | Definition of Similar Triangles | |||||||||||
\(\ds \) | \(=\) | \(\ds RQ\) | $OP$ is Radius of the Unit Circle | |||||||||||
\(\ds \) | \(=\) | \(\ds \tan \theta\) | Definition of Tangent Function |
When $\theta$ is obtuse, the same argument holds, but both $x$ and $\tan \theta$ are negative.
When $\theta = \dfrac \pi 2$ we have that $x = 0$.
Then $OP$ is parallel to the tangent line at $R$ which it therefore does not meet.
Thus when $\theta = \dfrac \pi 2$, it follows that $\tan \theta$ is not defined.
Likewise $\dfrac y x$ is not defined when $x = 0$.
Thus the relation holds for $\theta = \dfrac \pi 2$.
When $\pi < \theta < 2 \pi$ the diagram can be reflected in the $x$-axis.
In this case, $y$ is negative.
Thus the relation continues to hold.
When $\theta = 0$ and $\theta = \pi$ we have that $y = 0$ and $\tan \theta = 0 = \dfrac y x$.
Hence the result.
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.9$