How do we prove that two parametric equations are drawing the same thing? For example, if I have 
$$\begin {align}
x(t) &= r\sin t\cos t\\
y(t) &= r\sin^2 t\\
\end {align}$$ 
and 
$$\begin {align}
x(t) &= \frac r 2 \cos  t\\
y(t) &= \frac r 2 (\sin t + 1)
\end {align}$$
How do we show that the two parametric equations draw the same line?

Some efforts to establish that two parametrizations represent the same curve are trickier than others.  In your first set, you will need the "double-angle" trig identities, or their relatives (I am also going to write '$r$' as '$R$' , since it is intended to be a constant, rather than radius in polar coordinates):
$$\sin 2 \theta \ = \ 2 \sin\theta \cos\theta \ \rightarrow \ x(t) \ = \ R \ \sin t \cos t \ = \ R \cdot \frac{1}{2} \sin2t \ ,  $$
$$\sin^2 \theta \ = \ \frac{1}{2} (1 - \cos 2\theta) \ \rightarrow \ y(t) \ = \  R \  \sin^2 t  \ = \  R \cdot \frac{1}{2} (1- \cos 2t) $$
$$ R \ \sin 2t \ = \ 2x \ , \ 2y \ = \ R  -  R \ \cos 2t \ \Rightarrow \ R \ \cos 2t  \ = \ R  -  2y $$
$$ R^2 \sin^2 2t \ + \  R^2 \cos^2 2t \ = \ R^2 \ = \ (2x)^2 \ + \ ( R  -  2y)^2 \ = \ 4x^2  + \ R^2 \ - \ 4Ry \ +  4y^2  $$
$$ \Rightarrow \ 0 \  = \ 4x^2 \  - \ 4Ry \ + \ 4y^2 \ \Rightarrow \ x^2  \ + \ y^2 \  - \ Ry  \ = \ 0$$
$$  \Rightarrow \ x^2  \ + \ (y^2 \  - \ Ry \ + \frac{R^2}{4})  \ = \ \frac{R^2}{4} \ \Rightarrow \ x^2  \ + \ (y  -  \frac{R}{2})^2  \ = \ (\frac{R}{2})^2 \ ,$$
that is, a circle centered at $(0, \frac{R}{2})$ with radius $ \frac{R}{2}$ .  Since the trig functions both have a period of $\pi$, the parametrization starts on the circle with $t = 0$ at $(0,0)$, advances counter-clockwise to $(\frac{R}{2}, \frac{R}{2})$ at $t = \frac{\pi}{4}$, and continues around to return to the origin at $t = \pi$ .
With the second parametrization, we follow a similar process (so I won't show all the steps):
$$x(t) \ = \ \frac{R}{2} \cos t \  ,  \ y(t) \ = \  \frac{R}{2} ( \sin t \ + \ 1)$$
$$ R \ \cos t \ = \ 2x \ , \ 2y \ =   R \ \sin t \ + \ R \ \Rightarrow \ R \ \sin t  \ = \ R  -  2y $$
$$ R^2 \sin^2 t \ + \  R^2 \cos^2 t \ = \ R^2 \ = \ (2x)^2 \ + \ ( R  -  2y)^2  \ \Rightarrow \ x^2  \ + \ (y  -  \frac{R}{2})^2  \ = \ (\frac{R}{2})^2 \ .$$
So we have the same circle again.  However, this parametrization has a period of $2 \pi$ , starts at $(\frac{R}{2}, \frac{R}{2})$ for $t = 0$ , advances counter-clockwise to $(0,R)$ at $t = \frac{\pi}{2}$ , and completes its cycle at $(\frac{R}{2}, \frac{R}{2})$ for $t = 2 \pi$ .  (As mjqxxx has said, the circular trajectory is followed in the same direction, but at different "speeds" and "starting points".)
You may well imagine that an unlimited number of other parametrizations can be chosen, differing not only in "rate of travel", "direction of travel", and "initial point", but also using "speed functions" which are not constant (or even reverse themselves!).  Because this is possible, there are many ways to "disguise" a parametric curve; consequently, there is no general method for checking whether two descriptions actually produce the same figure.
