Distance between Two Points in Plane in Polar Coordinates/Proof 2
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Theorem
Let $A = \polar {r_1, \theta_1}$ and $B = \polar {r_2, \theta_2}$ be two points in a polar coordinate plane
The distance $d$ between $A$ and $B$ is given by:
- $d = \sqrt {r_1^2 + r_2^2 + 2 r_1 r_2 \map \cos {\theta_1 - \theta_2} }$
Proof
From Conversion between Cartesian and Polar Coordinates in Plane, $A$ and $B$ can be expressed in Cartesian coordinates as follows:
| \(\ds x_1\) | \(=\) | \(\ds r_1 \cos \theta_1\) | ||||||||||||
| \(\ds y_1\) | \(=\) | \(\ds r_1 \sin \theta_1\) | ||||||||||||
| \(\ds x_2\) | \(=\) | \(\ds r_2 \cos \theta_2\) | ||||||||||||
| \(\ds y_2\) | \(=\) | \(\ds r_2 \sin \theta_2\) |
Thus the distance $d$ between $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ is given by:
| \(\ds d^2\) | \(=\) | \(\ds \paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2\) | Distance Formula | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {r_1 \cos \theta_1 - r_2 \cos \theta_2}^2 + \paren {r_1 \sin \theta_1 - r_2 \sin \theta_2}^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds r_1^2 \paren {\cos^2 \theta_1 + \sin^2 \theta_1} + r_2^2 \paren {\cos^2 \theta_2 + \sin^2 \theta_2} + 2 r_1 r_2 \paren {\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds r_1^2 + r_2^2 + 2 r_1 r_2 \paren {\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2}\) | Sum of Squares of Sine and Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds r_1^2 + r_2^2 + 2 r_1 r_2 \map \cos {\theta_1 - \theta_2}\) | Cosine of Difference |
The result follows.
$\blacksquare$