Sine 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:
- $\sin \theta = \dfrac y r$
where $\sin$ denotes the sine 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 $OP$.
$\angle OSP = \angle ORQ$, 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.
By definition of similarity:
Then:
\(\ds \frac y r\) | \(=\) | \(\ds \frac {SP} {OP}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {RQ} {OQ}\) | Definition of Similar Triangles | |||||||||||
\(\ds \) | \(=\) | \(\ds RQ\) | $OP$ is Radius of Unit Circle | |||||||||||
\(\ds \) | \(=\) | \(\ds \sin \theta\) | Definition of Sine |
When $\theta$ is obtuse, the same argument holds.
When $\theta = \dfrac \pi 2$ we have that $x = 0$.
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.
In this case, both $\sin \theta$ and $y$ are negative.
Thus the relation continues to hold.
When $\theta = 0$ and $\theta = \pi$ we have that $y = 0$ and $\sin \theta = 0 = \dfrac y r$.
Hence the result.
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.7$