Example 7. Write the equation of the evolute of the cycloid

$$
x=a(t-\sin t), y=a(1-\cos t)
$$
Solution. Since

$$
x^{\prime 2}+y^{\prime 2}=2 a^{2}(1-\cos t), x^{\prime} y^{\prime \prime}-y^{\prime} x^{\prime \prime}=-a^{2}(1-\cos t)
$$

(see example 5), then according to formulas (3.7) we get

$$
\begin{gathered}
X=x-y^{\prime} \cdot \frac{x^{\prime 2}+y^{\prime 2}}{x^{\prime} y^{\prime \prime}-y^{\prime} x^{\prime \prime}}=a(t-\sin t)-a(\sin t) \cdot \frac{2 a^{2}(1-\cos t)}{-a^{2}(1-\cos t)}= \\
=a t-a \sin t+2 a \sin t=a t+a \sin t \\
Y=y+x^{\prime} \cdot \frac{x^{\prime 2}+y^{\prime 2}}{x^{\prime} y^{\prime \prime}-y^{\prime} x^{\prime \prime}}=a(1-\cos t)+a(1- \\
-\cos t) \cdot \frac{2 a^{2}(1-\cos t)}{-a^{2}(1-\cos t)}=a(1-\cos t)-2 a(1-\cos t)= \\
=-a(1-\cos t)
\end{gathered}
$$

Therefore, the equations of the evolute of the cycloid are:

$$
X=a t+a \sin t ; Y=-a(1-\cos t)
$$

Introduce a new variable by the formula $t=\pi+\tau$. The equations of the evolute will take the form:

$$
\begin{aligned}
X= & a[\pi+\tau+\sin (\pi+\tau)]=a \pi+a \tau-a \sin \tau=a(\tau-\sin \tau)+a \pi \\
& Y=-a[1-\cos (\pi+\tau)]=-a(1+\cos \tau)=-a-a \cos \tau= \\
= & -2 a+a-a \cos \tau=a-a \cos \tau-2 a=a(1-\cos \tau)-2 a
\end{aligned}
$$

That is,

$$
X=a(\tau-\sin \tau)+a \pi ; \quad Y=a(1-\cos \tau)-2 a
$$

These equations can be written as:

$$
x=a\left(t^{\prime}-\sin t^{\prime}\right) ; y=a\left(1-\cos t^{\prime}\right)
$$

where $x=X-a \pi, y=Y+2 a, \tau=t^{\prime}$.

Therefore, the evolute of the cycloid is the same cycloid, shifted along the $O x$ axis by the amount $\pi a$ and along the $O y$ axis by the amount $2 a$ (Fig. 3.3).