Theorem: Length of Arc of Deltoid

The total length of the arcs of a deltoid constructed within a deferent of radius $a$ is given by:
:$\LL = \dfrac {16 a} 3$

Proof:
Let $H$ be embedded in a cartesian plane with its center at the origin and one of its cusps positioned at $\tuple {a, 0}$.
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We have that $\LL$ is $3$ times the length of one arc of the deltoid.
From Arc Length for Parametric Equations:
:$\ds \LL = 3 \int_{\theta \mathop = 0}^{\theta \mathop = 2 \pi/3} \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$
where, from Equation of Deltoid:
:$\begin{cases}
x & = 2 b \cos \theta + b \cos 2 \theta \\
y & = 2 b \sin \theta - b \sin 2 \theta
\end{cases}$
We have:
{{begin-eqn}}
{{eqn | l = \frac {\d x} {\d \theta}
      | r = -2 b \sin \theta - 2 b \sin 2 \theta
      | c = 
}}
{{eqn | l = \frac {\d y} {\d \theta}
      | r = 2 b \cos \theta - 2 b \cos 2 \theta
      | c = 
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | o = 
      | r = \paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2
      | c = 
}}
{{eqn | r = \paren {-2 b \sin \theta - 2 b \sin 2 \theta}^2 + \paren {2 b \cos \theta - 2 b \cos 2 \theta}^2
      | c = 
}}
{{eqn | r = 4 b^2 \paren {\paren {-\sin \theta - \sin 2 \theta}^2 + \paren {\cos \theta - \cos 2 \theta}^2}
      | c = 
}}
{{eqn | r = 4 b^2 \paren {\sin^2 \theta + 2 \sin \theta \sin 2 \theta + \sin^2 2 \theta + \cos^2 \theta - 2 \cos \theta \cos 2 \theta + \cos^2 2 \theta}
      | c = 
}}
{{eqn | r = 4 b^2 \paren {2 + 2 \sin \theta \sin 2 \theta - 2 \cos \theta \cos 2 \theta}
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 8 b^2 \paren {1 + \sin \theta \sin 2 \theta - \cos \theta \cos 2 \theta}
      | c = 
}}
{{eqn | r = 8 b^2 \paren {1 + 2 \sin^2 \theta \cos \theta - \cos \theta \cos 2 \theta}
      | c = Double Angle Formula for Sine
}}
{{eqn | r = 8 b^2 \paren {1 + 2 \sin^2 \theta \cos \theta - \cos \theta \paren {1 - 2 \sin^2 \theta} }
      | c = {{Corollary|Double Angle Formula for Cosine|2}}
}}
{{eqn | r = 8 b^2 \paren {1 - \cos \theta + 4 \sin^2 \theta \cos \theta}
      | c = simplifying
}}
{{eqn | r = 8 b^2 \paren {1 - \cos \theta + 4 \cos \theta \paren {1 - \cos^2 \theta} }
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 8 b^2 \paren {1 - \cos \theta + 4 \cos \theta \paren {1 + \cos \theta} \paren {1 - \cos \theta} }
      | c = Difference of Two Squares
}}
{{eqn | r = 8 b^2 \paren {1 - \cos \theta} \paren {1 + 4 \cos \theta \paren {1 + \cos \theta} }
      | c = 
}}
{{eqn | r = 8 b^2 \paren {1 - \cos \theta} \paren {1 + 4 \cos \theta + 4 \cos^2 \theta}
      | c = 
}}
{{eqn | r = 8 b^2 \paren {1 - \cos \theta} \paren {1 + 2 \cos \theta}^2
      | c = 
}}
{{eqn | r = 8 b^2 \paren {2 \sin^2 \frac \theta 2} \paren {1 + 2 \cos \theta}^2
      | c = Half Angle Formula for Sine
}}
{{eqn | r = 16 b^2 \sin^2 \frac \theta 2 \paren {1 + 2 \cos \theta}^2
      | c = 
}}
{{end-eqn}}
Thus:
:$\sqrt {\paren {\dfrac {\d x} {\d \theta} }^2 + \paren {\dfrac {\d y} {\d \theta} }^2} = 4 b \sin \dfrac \theta 2 \size {1 + 2 \cos \theta}$
In the range $0$ to $2 \pi / 3$, $1 + 2 \cos \theta$ is not less than $0$, and so:
:$\ds \LL = 3 \int_0^{2 \pi / 3} 4 b \sin \dfrac \theta 2 \paren {1 + 2 \cos \theta} \rd \theta$
Put:
:$u = \cos \dfrac \theta 2$
so:
:$2 \dfrac {\d u} {\d \theta} = -\sin \dfrac \theta 2$
As $\theta$ increases from $0$ to $\dfrac {2 \pi} 3$, $u$ decreases from $1$ to $\dfrac 1 2$.
Then:
{{begin-eqn}}
{{eqn | l = 1 + 2 \cos \theta
      | r = 1 + 2 \paren {2 \paren {\cos \dfrac \theta 2}^2 - 1}
      | c = Half Angle Formula for Cosine
}}
{{eqn | r = 4 u^2 - 1
      | c = 
}}
{{end-eqn}}
Substituting:
:$2 \dfrac {\d u} {\d \theta} = -\sin \dfrac \theta 2$
and the limits of integration:
:$u = 1$ for $\theta = 0$
:$u = \dfrac 1 2$ for $\theta = \dfrac {2 \pi} 3$
we obtain, after simplifying the sign:
{{begin-eqn}}
{{eqn | l = \LL
      | r = 12 b \int_1^{1/2} \paren {1 - 4 u^2} 2 \rd u
      | c = Integration by Substitution
}}
{{eqn | r = 24 b \intlimits {u - \frac 4 3 u^3} 1 {1/2}
      | c = 
}}
{{eqn | r = 24 b \paren {\paren {\frac 1 2 - \frac 4 3 \frac 1 {2^3} } - \paren {1 - \frac 4 3} }
      | c = 
}}
{{eqn | r = 24 b \paren {\frac 1 2 - \frac 1 6 - 1 + \frac 4 3}
      | c = 
}}
{{eqn | r = 24 b \paren {\frac 1 3 + \frac 1 3}
      | c = 
}}
{{eqn | r = 16 b
      | c = 
}}
{{eqn | r = \frac {16 a} 3
      | c = 
}}
{{end-eqn}}
{{qed}}