Theorem: Length of Arc of Cycloid

Let $C$ be a cycloid generated by the equations:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
Then the length of one arc of the cycloid is $8 a$.

Proof:
{{AuthorRef|René Descartes|Descartes}} approximated the cycloid by substituting a polygon for the generating circle.  
For example, a hexagon (where $n = 6$) produces $5$ arches (that is, $n - 1$).  
Each arch is generated by turning through the external angle of the polygon.  
For the hexagon, this is $\dfrac {2 \pi} 6$.
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The arches have arms of different lengths.
The first is simply the side of the hexagon.  
The arm of the second arch is the chord which forms a triangle with two adjacent sides of the hexagon.  
The arm of the third arch is the chord which forms a trapezium with three adjacent sides of the hexagon, etc.  
We can obtain the lengths of those sides by considering the angles that subtend them.  
The central angle subtending a single side of the hexagon is:
{{begin-eqn}}
{{eqn | l = \theta
      | r = \frac {2 \pi} 6
      | c = central angle for hexagon
}}
{{eqn | l = \theta
      | r = k \frac {2 \pi} 6
      | c = angle subtending k sides
}}
{{eqn | l = c
      | r = 2 a \map \sin {k \times \frac 1 2 \times \frac {2 \pi} 6}
      | c = chord for hexagon '''inscribed''' in circle of radius a, subtending k sides and Inscribed Angle Theorem
}}
{{eqn | l = l
      | r = \frac {2 \pi} 6 \cdot 2 a \map \sin {\frac {k \pi} 6}
      | c = Arc Length of Sector
}}
{{eqn | l = L
      | r = \sum_{k \mathop = 0}^6 \frac {2 \pi} 6 \cdot 2 a \map \sin {\frac  {k \pi} 6}
      | c = total arc, summed over $k = n - 1$ arches
}}
{{end-eqn}}
Let the sum range from $0$ to $6$ (the two extra terms contribute nothing).
In addition, everywhere $6$ appears we can substitute $n$ for the polygon of $n$ sides.
{{begin-eqn}}
{{eqn | l = L
      | r = \sum_{k \mathop = 0}^n \frac {2 \pi} n \cdot 2 a \cdot \map \sin {\frac {k \pi} n }
      | c = 
}}
{{eqn | r = 4 a \frac \pi n \sum_{k \mathop = 0}^n \map \sin {k \theta}
      | c = 
}}
{{end-eqn}}
where $\theta = \dfrac \pi n$.  
From Lagrange's Sine Identity, we have:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \map \sin {k \theta}
      | r = \frac {\map \cos {\frac 1 2 \theta} - \map \cos {n \theta + \frac 1 2 \theta} } {2 \map \sin {\frac 1 2 \theta} }
      | c = 
}}
{{eqn | ll = \leadsto
      | l = \sum_{k \mathop = 0}^n \map \sin {\frac {k \pi} n}
      | r = \frac {\map \cos {\frac \pi {2 n} } - \map \cos {\pi + \frac \pi {2 n} } } {2 \map \sin {\frac \pi {2 n} } }
      | c = $\theta = \dfrac \pi n$
}}
{{end-eqn}}
Now let $n \rightarrow \infty$.  
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^\infty \map \sin {\frac {k \pi} n}
      | r = \lim_{n \mathop \to \infty} \frac {\map \cos {\frac \pi {2 n} } - \map \cos {\pi + \frac \pi {2 n} } } {2 \map \sin {\frac \pi {2 n} } }
      | c = 
}}
{{eqn | r = \frac {\map \cos 0 + \map \cos 0} {2 \map \sin {\frac \pi {2 n} } }
      | c = Cosine of Angle plus Straight Angle 
}}
{{eqn | r = \frac 2 {2 \times \frac \pi {2 n } }
      | c = {{cos|0}} and Limit of $\dfrac {\sin x} x$ at $0$
}}
{{eqn | r = \frac {2 n} \pi
      | c = 
}}
{{end-eqn}} 
Therefore, the expression for the total arc length $L$ becomes:
{{begin-eqn}}
{{eqn | l = L
      | r = 4a \cdot \dfrac \pi n \dfrac {2n} \pi = 8a
      | c = 
}}
{{end-eqn}}
{{qed}}