# Cosine 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:

- $\cos \theta = \dfrac x r$

where $\cos$ denotes the cosine of $\theta$.

## Proof

Let a unit circle $C$ be drawn with its center at the origin $O$.

Let $Q$ be the point on $C$ which intersects $OP$.

From Parallelism implies Equal Alternate Angles, $\angle OQR = \theta$.

Thus:

- $(1): \quad \cos \theta = RQ$

by definition of cosine

$\angle OSP = \angle ORQ$, as both are right angles.

Both $\triangle OSP$ and $\triangle ORQ$ share angle $\angle SOP$.

By Triangles with Two Equal Angles are Similar it follows that $\triangle OSP$ and $\triangle ORQ$ are similar.

By definition of similarity:

Then:

\(\displaystyle \frac x r\) | \(=\) | \(\displaystyle \frac {SP} {OP}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {RQ} {OQ}\) | Definition of Similar Triangles | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle RQ\) | as $OP$ is the radius of the unit circle | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \cos \theta\) | from $(1)$ above |

$\blacksquare$

When $\theta$ is obtuse, the same argument holds, except that this time both $x$ and $\cos \theta$ are negative.

When $\theta = \dfrac \pi 2$ we have that $x = 0$ and $\cos \theta = 0 = \dfrac x r$.

Thus $y = r$ and $\sin \theta = 1 \dfrac y r$.

Thus the relation holds for $\theta = \dfrac \pi 2$.

When $\pi < \theta < 2 \pi$ the diagram can be reflected in the $x$-axis.

Thus the relation continues to hold.

When $\theta = 0$ we have that $y = 0$ and $\cos \theta = 1 = \dfrac x r$.

When $\theta = \pi$ we have that $y = 0$ and $x = -r$, while $\cos \theta = -1 = \dfrac x r$.

Hence the result.

$\blacksquare$

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.8$