Relation between Equations for Hypocycloid and Epicycloid

Theorem

Consider the hypocycloid defined by the equations:

$x = \paren {a - b} \cos \theta + b \map \cos {\paren {\dfrac {a - b} b} \theta}$

$y = \paren {a - b} \sin \theta - b \map \sin {\paren {\dfrac {a - b} b} \theta}$

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

$x = \paren {a + b} \cos \theta - b \map \cos {\paren {\dfrac {a + b} b} \theta}$

$y = \paren {a + b} \sin \theta - b \map \sin {\paren {\dfrac {a + b} b} \theta}$

Proof

\(\ds x\)

\(=\)

\(\ds \paren {a - \paren {-b} } \cos \theta + \paren {-b} \map \cos {\paren {\dfrac {a - \paren {-b} } {\paren {-b} } } \theta}\)

putting $-b$ for $b$

\(\ds \)

\(=\)

\(\ds \paren {a + b} \cos \theta - b \map \cos {-\paren {\dfrac {a + b} b} \theta}\)

\(\ds \)

\(=\)

\(\ds \paren {a + b} \cos \theta - b \map \cos {\paren {\dfrac {a + b} b} \theta}\)

Cosine Function is Even

\(\ds y\)

\(=\)

\(\ds \paren {a - \paren {-b} } \sin \theta - \paren {-b} \map \sin {\paren {\dfrac {a - \paren {-b} } {\paren {-b} } } \theta}\)

putting $-b$ for $b$

\(\ds \)

\(=\)

\(\ds \paren {a + b} \sin \theta + b \map \sin {-\paren {\dfrac {a + b} b} \theta}\)

\(\ds \)

\(=\)

\(\ds \paren {a + b} \sin \theta - b \map \sin {\paren {\dfrac {a + b} b} \theta}\)

Sine Function is Odd

$\blacksquare$

Sources

• 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.21$: The Cycloid: Problem $13$