Canonical to Parametric, Ellipse Equation I've done some algebra tricks in this derivation and I'm not sure if it's okay to do those things.
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = \cos^2\theta + \sin^2\theta$$
Can I really do this next step? 
$$\frac{x^2}{a^2} = \cos^2\theta\quad\text{and}\quad\frac{y^2}{b^2} = \sin^2\theta$$
$$x^2 = a^2\cos^2\theta\quad\text{and}\quad y^2 = b^2\sin^2\theta$$
Ignoring the negative numbers:
$$x = a\cos\theta\quad\text{and}\quad y = b\sin\theta$$

The idea behind your argument is absolutely fine.  Any two non-negative numbers $u$ and $v$ such that $u+v=1$ can be expressed as $u=\cos^2\theta$, $v=\sin^2\theta$ for some $\theta$. This is so obvious that it probably does not require proof. Set $u=\cos^2\theta$. Then $v=1-\cos^2\theta=\sin^2\theta$.   
The second displayed formula muddies things somewhat. You intended to say that if $x^2/a^2+y^2/b^2=1$, then there exists a $\theta$ such that $x^2/a^2=\cos^2\theta$ and $y^2/b^2=\sin^2\theta$.  You did not mean that for any $\theta$, if $x^2/a^2+y^2/b^2=1$ then $x^2/a^2=\cos^2\theta$!  But the transition from the second displayed equation to the third could be interpreted as asserting what you clearly did not intend to say.  
It would be better to do exactly what you did, but to use more geometric language, as follows. 
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad\text{iff}\quad \left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2=1.$$ 
But the equation on the right holds iff the point $(x/a, y/b)$ lies on the unit circle.  The points on the unit circle are parametrized by $(\cos \theta,\sin\theta)$, with $\theta$ ranging over $[0,2\pi)$, so the points on our ellipse are given by $x=a\cos\theta$, $y=a\sin\theta$.
