# Slope of Tangent to Cycloid/Proof 1

## Theorem

Let $C$ be a cycloid generated by the equations:

$x = a \paren {\theta - \sin \theta}$
$y = a \paren {1 - \cos \theta}$

The slope of the tangent to $C$ at the point $\tuple {x, y}$ is given by:

$\dfrac {\d y} {\d x} = \cot \dfrac \theta 2$

## Proof

By definition of Derivative of Curve at Point, the tangent to $C$ at the point $\tuple {x, y}$ is the derivative of its equation at that point.

Thus:

 $\ds \frac {\d x} {\d \theta}$ $=$ $\ds a \paren {1 - \cos \theta}$ $\ds \frac {\d y} {\d \theta}$ $=$ $\ds a \sin \theta$ $\ds \leadsto \ \$ $\ds \frac {\d y} {\d x}$ $=$ $\ds \frac {\d y} {\d \theta} / \frac {\d x} {\d \theta}$ $\ds$ $=$ $\ds \frac {a \sin \theta} {a \paren {1 - \cos \theta} }$ $\ds$ $=$ $\ds \frac {\sin \theta} {1 - \cos \theta}$ $\ds$ $=$ $\ds \frac 1 {\map \tan {\theta / 2} }$ Half Angle Formula for Tangent: Corollary 2 $\ds$ $=$ $\ds \cot \frac \theta 2$

$\blacksquare$