$\begin{gathered}x=a(t+\sin t) \text { and } y=a(1-\cos t) \\ \frac{d x}{d t}=a(1+\cos t), \frac{d y}{d t}=a \sin t \\ \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a \sin t}{a(1+\cos t)}=\frac{2 \sin t / 2 \cos t / 2}{2 \cos ^2 t / 2} \\ \frac{d y}{d x}=\tan \frac{t}{2} \\ \frac{d^2 y}{d x^2}=\sec ^2 t / 2 \cdot \frac{1}{2} \cdot \frac{d t}{d x} \\ =\frac{\sec ^2 \frac{t}{2}}{2 a(1+\cos t)}=\frac{1}{4 a \cos ^4(t / 2)} .\end{gathered}$