How to prove that the parametric representation $(\cos t,\sin t)$ is indeed the unit circle? Since the implicit equation of unit circle is $x^2+y^2=1$, the parametric representation $x=\cos t, y=\sin t$ satisfies the equation. At the same time, both equation have the same domain, therefore I  can conclude that the parametric representation is at least part of the unit circle. However, how do we prove that it is indeed the unit circle? Is the parametric representation unique for every curve? How do we find the parametric representation of a curve in general?
 A: Take a point $P$ is on the circle at a fixed distance $r$ (the radius) from the center. The point $P$ subtends an angle $\theta $ to the positive x-axis.
Using trigonometry, we can find the coordinates of $P$ from the right triangle shown. In this triangle the radius $r$ is the hypotenuse $h $ of the triangle.
(source: mathopenref.com)
We can easily see that $\sin \theta =\frac {o}{h} $. But here $o $ is equal to the ordinate $y $ of the point $P$ and we know that $h=r$. So thus, we have, $$y=r\sin \theta$$ By similar arguments we can also show that $x=r \cos \theta $ giving us the required parametrization.

For an alternative way, see here. Hope it helps.
A: The easy way to see this is to remember that the point on the unit circle at angle $\theta$ from the positive x axis is given by $x=\cos(\theta),$ $y = \sin(\theta).$ Thus when $t$ runs from $0$ to $2\pi,$ you sweep the unit circle exactly once.
The parametric representation isn't unique. For instance $x = \sin(t)$, $y = \cos(t)$ would work equally well. it would just sweep the unit circle clockwise starting from $(0,1)$ rather than counterclockwise from $(1,0).$ 
For a less trivial example, take the piecewise
$$
f(t) = \left\{\begin{array}{llc} x = t-1,& y= \sqrt{1-(t-1)^2} & 0 \le t<2\\ x = 3-t,&y=-\sqrt{1-(3-t)^2}&  2\le t < 4 \end{array}\right.
$$
which first sweeps out the upper semicircle, and then the lower semicircle in a clockwise fashion. 
How to find a parametric representation in general for any curve is a bit of a tough question, to which I know no simple answer.
A: When $-{\pi\over2}<t<{\pi\over2}$ then $\cos t>0$ and $t\mapsto\sin t$ is  strictly increasing from $-1$ to $1$. This and continuity allow to conclude that
$$t\mapsto(\cos t,\sin t)\qquad\left(-{\pi\over2}<t<{\pi\over2}\right)$$
produces the right half of the unit circle (without the endpoints) in a one-one way. Argue similarly for the $t$-intervals $\>]0,\pi[\>$, $\bigl]{\pi\over2},{3\pi\over2}\bigr[\>$, $\>]\pi,2\pi[\>$ to convince yourself that $t\mapsto(\cos t,\sin t)$ indeed produces the full circle.
Concerning parametric representations in general: A curve is a static object, drawn on a piece of paper. Contrasting this a parametric representation is an exact timetable for the "pencil" (or "moving point") drawing this curve, hence carries information which maybe you are not interested in. On the other hand in order to compute various geometric quantities related to this curve, like length or enclosed area, a parametric representation is indispensable. Who would guess that the innocent looking equation $x^2+y^2=1$ is intimately related to the number $\pi$?
