Equation of Epicycloid

Theorem
Let a circle $C_1$ of radius $b$ roll without slipping around the outside of a circle $C_2$ of radius $a$.

Let $C_2$ be embedded in a cartesian coordinate plane with its center $O$ located at the origin.

Let $P$ be a point on the circumference of $C_1$.

Let $C_1$ be initially positioned so that $P$ is its point of tangency to $C_2$, located at point $A = \left({a, 0}\right)$ on the $x$-axis.

Let $H$ be the epicycloid traced out by the point $P$.

Let $\left({x, y}\right)$ be the coordinates of $P$ as it travels over the plane.

The point $P = \left({x, y}\right)$ is described by the equations:
 * $x = \left({a + b}\right) \cos \theta - b \cos \left({\left({\dfrac {a + b} b}\right) \theta}\right)$
 * $y = \left({a + b}\right) \sin \theta - b \sin \left({\left({\dfrac {a + b} b}\right) \theta}\right)$

Proof

 * Epicycloid.png

Let $C_1$ have rolled so that the line $OC$ through the radii of $C_1$ and $C_2$ is at angle $\theta$ to the $x$-axis.

Let $C_1$ have turned through an angle $\phi$ to reach that point.

By definition of sine and cosine, $P = \left({x, y}\right)$ is defined by:
 * $x = \left({a + b}\right) \cos \theta - b \cos \left({\phi + \theta}\right)$
 * $y = \left({a + b}\right) \sin \theta - b \sin \left({\phi + \theta}\right)$

The arc of $C_1$ between $P$ and $B$ is the same as the arc of $C_2$ between $A$ and $B$.

Thus by Arc Length of Sector:
 * $ a \theta = b \phi$

Thus:
 * $\phi + \theta = \left({\dfrac {a + b} b}\right) \theta$

whence the result.