x = t − sin(t)

y = 1 − cos(t).

This comes from imagining a circle whose radius is 1 and whose center is at <t, 1> at time t, rolling along the line y = 0. P is the point on the circumference of the circle leaving the trace.

A half height circle rolling at the same velocity and turning twice as fast yields the half cycloid:

x = t − sin(2t)/2

y = ½ − cos(2t)/2.

The small circle can be seen as rolling inside the large circle
— for all values of t they are mutually tangent at <t, 0>.
The point p is on the small circle that starts at <0, 0> and traverses the small cycloid.
p is always on the diameter of the large circle thru P.

p = < t − sin(2t)/2, ½ − cos(2t)/2>

See this lovely animation.

P moves as in the first equations.
The other end P' of the diameter moves thus:

P' = < t + sin(t), 1 + cos(t)>

p moves harmonically between P and P' thus:

p = ((1 + cos(t))P + (1 − cos(t))P')/2

=<(1 + cos(t))(t − sin(t)) + (1 − cos(t))( t + sin(t)),
(1 + cos(t))(1 − cos(t)) + (1 − cos(t))(1 + cos(t))>/2

= < t − cos(t)sin(t), 1 − cos^{2}(t)>
= < t − sin(2t)/2, ½ − cos(2t)/2>

which agrees with the description of p as a point on the small circle.

Now if p's velocity is always parallel with the large diameter
then p will be on the envelope of that diameter.

dp/dt = d< t − sin(2t)/2, ½ − cos(2t)/2>/dt

= < d(2t − sin(2t))/dt, d(1 − cos(2t))/dt>/2
= < 2 − 2cos(2t), 2sin(2t)>/2
= <1 − cos(2t), sin(2t)>.

The diameter as vector is <cos(t), sin(t)> which is parallel to <cos(t), sin(t)>sin(t)
= <cos(t)sin(t), sin^{2}(t)>

The claim is that the diameter of the large circle thru P is always tangent to the small cycloid. Thus the small cycloid is the envelope of the diameter of the big circle.