Example 3. Form the equations of the tangent and normal to the curve given parametrically,

$$
x=2(t-\sin t) ; y=2(1-\cos t)
$$

1) at an arbitrary point; 2) at $t=\frac{\pi}{2}$.
Solution. The equation of the tangent (3.1) can be written as

$$
Y-y=y_{x}^{\prime}(x)(X-x)
$$

where $(x, y)$ are the coordinates of a point on the curve, and $(X, Y)$ are the coordinates of a point on the tangent.

We find the derivative $y_{x}^{\prime}$ using the formula

$$
y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{2 \sin t}{2(1-\cos t)}=\frac{\sin t}{1-\cos t}=\operatorname{ctg} \frac{t}{2}
$$

Substituting the expressions for $x, y$ and $y_{x}^{\prime}$ into formula (3.1'), we obtain the equation of the tangent to the given curve at any point

$$
Y-2(1-\cos t)=\operatorname{ctg} \frac{t}{2}[X-2(t-\sin t)]
$$

For $t=\frac{\pi}{2}$, from the last equation we find

$$
Y-2=1 \cdot\left[X-2\left(\frac{\pi}{2}-1\right)\right], Y-2=X-\pi+2
$$

or

$$
X-Y-\pi+4=0
$$