Relation between Equations for Hypocycloid and Epicycloid

Theorem

Consider the hypocycloid defined by the equations:

$x = \left({a - b}\right) \cos \theta + b \cos \left({\left({\dfrac {a - b} b}\right) \theta}\right)$
$y = \left({a - b}\right) \sin \theta - b \sin \left({\left({\dfrac {a - b} b}\right) \theta}\right)$

By replacing $b$ with $-b$, this converts to the equations which define an epicycloid:

$x = \left({a + b}\right) \cos \theta - b \cos \left({\left({\dfrac {a + b} b}\right) \theta}\right)$
$y = \left({a + b}\right) \sin \theta - b \sin \left({\left({\dfrac {a + b} b}\right) \theta}\right)$

Proof

 $\displaystyle x$ $=$ $\displaystyle \left({a - \left({-b}\right)}\right) \cos \theta + \left({-b}\right) \cos \left({\left({\dfrac {a - \left({-b}\right)} {\left({-b}\right)} }\right) \theta}\right)$ putting $-b$ for $b$ $\displaystyle$ $=$ $\displaystyle \left({a + b}\right) \cos \theta - b \cos \left({-\left({\dfrac {a + b} b}\right) \theta}\right)$ $\displaystyle$ $=$ $\displaystyle \left({a + b}\right) \cos \theta - b \cos \left({\left({\dfrac {a + b} b}\right) \theta}\right)$ Cosine Function is Even

 $\displaystyle y$ $=$ $\displaystyle \left({a - \left({-b}\right)}\right) \sin \theta - \left({-b}\right) \sin \left({\left({\dfrac {a - \left({-b}\right)} {\left({-b}\right)} }\right) \theta}\right)$ putting $-b$ for $b$ $\displaystyle$ $=$ $\displaystyle \left({a + b}\right) \sin \theta + b \sin \left({-\left({\dfrac {a + b} b}\right) \theta}\right)$ $\displaystyle$ $=$ $\displaystyle \left({a + b}\right) \sin \theta - b \sin \left({\left({\dfrac {a + b} b}\right) \theta}\right)$ Sine Function is Odd

$\blacksquare$