Geometric Proof for the Derivative of Sine I was following a geometric proof for the derivative of sine when I came across this unjustified assertion: In the diagram, as we choose Q ever closer to P, the chord PQ approximates the corresponding arc arbitrarily well. While I understand the intuitive reason for this, how could this idea be made more rigorous so that the proof is guaranteed to work. Something with epsilon delta was what I was thinking?
 

The comments that have been (thus far) made on the OP have completely missed the issue, which is how do you know that the length of $\overline{PQ}$ approximates the arclength of $\widehat{PQ}$?
Unfortunately, that difficultly is intractable at the level where this proof is given. With rare exceptions, students at this level have never seen a definition for arclength. It is (like many other notions) just a nebulous concept that everyone understands and assumes exists, but no one has a rigorous grounding in.
Rather than go into the question of how arclength is rigorously defined, though, I suggest a different proof. It also depends on a not-yet-rigorously defined concept, Area, but one for which a useful mathematical relationship is known: If $A \subseteq B$, then $\operatorname{Area}(A) \le \operatorname{Area}(B)$.
The area of a sector of a circle of radius $r$ subtending an angle $\theta$ (in radians) is $\frac{\theta r^2}2$, a formula which can be proven at this level fairly easily, given that the area of the entire circle is $\pi r^2$. 
In this diagram

The inner sector has area $\frac{\theta \cdot 1^2}2 = \frac\theta2$. The outer circle has radius $\sqrt{1+\sin^2 \theta}$, so the outer sector has area $\frac{\theta(1 + \sin^2 \theta)}2$. The triangle has base $1$ and height $\left(\sqrt{1+\sin^2\theta}\right)\sin \theta$,so its area is $\frac 12\left(\sqrt{1+\sin^2\theta}\right)\sin\theta$. As the inner sector is contained in the triangle, which is contained in the outer sector, we have
$$\theta \le \left(\sqrt{1+\sin^2\theta}\right)\sin\theta \le \theta(1 + \sin^2 \theta)\\\frac \theta{\sqrt 2}\le\frac \theta{\sqrt{1+\sin^2\theta}} \le \sin \theta \le \theta\sqrt{1+\sin^2\theta}\le\sqrt2 \theta$$
Which gives $$\lim_{\theta \to 0}~ \sin \theta = 0$$ And dividing by $\theta$, we get
$$\frac 1{\sqrt{1+\sin^2\theta}} \le \frac {\sin \theta}\theta \le \sqrt{1+\sin^2\theta}$$
which gives $$\lim_{\theta \to 0} \frac{\sin \theta}\theta = 1$$
