Can you show how triangles are related to harmonic growth without using sine? Here is my predicament: I understand how right triangles can be inscribed within a unit circle in such a way that it becomes obvious that sine = opposite and cosine = adjacent (w.r.t. the angle between the hypotenuse and the x axis). I believe this, but I don't understand why it is the case.
What it breaks down to (in my mind) is this:
If you vary an angle (theta) of a right triangle linearly from zero,
then the height of the triangle is harmonic (plots out sine).

In more concrete terms
You have a pole lying on the ground. You slowly lift the pole, so that
the angle between pole and ground increases linearly. At what rate
does the height of the tip of the pole increase?

I understand quite well that the answer to the above is $\sin(\theta)$, but I don't understand why that is the case. Can you prove either of the above statements without using $sine$ or the unit circle?
I can imagine proofs where, for example, you prove that because you are increasing the angle linearly the tip of the pole rises vertically with acceleration inversely proportional to it's height (i.e. utilizing the $y''=-y$ definition of harmonics), but I can't construct such a proof without presupposing the behavior of sine.
That's the sort of proof I'm looking for. Can you prove, without sine, that the height of a right triangle (as you increas the $\theta$ linearly) traces out a simple harmonic motion?
 A: Let the base of the pole be at the origin, $(0,0)$, and let the tip of the pole when tilted an angle $\theta$ be at the point $(x(\theta),y(\theta))$.  The distance to the origin is always constant, and the speed of the tip is constant if $\theta$ increases at a constant rate.  Thus, 
$$(x(\theta))^2+(y(\theta))^2=\text{constant},$$
and
$$(x'(\theta))^2+(y'(\theta))^2=\text{constant}.$$
Differentiating each of these equations and dividing by $2$ yields
$$x(\theta)x'(\theta)+y(\theta)y'(\theta) = 0,$$
and
$$x'(\theta)x''(\theta)+y'(\theta)y''(\theta)=0.$$
This implies that $(x''(\theta),y''(\theta))$ lies on the same line through the origin as $(x(\theta),y(\theta))$, perpendicular to the line through the origin and $(x'(\theta),y'(\theta))$.  It follows that $x''(\theta) =K(\theta)x(\theta)$ and $y''(\theta)=K(\theta)y(\theta)$ for some function $K$.  Differentiating $xx'+yy'=0$ again yields
$$(x'(\theta))^2+(y'(\theta))^2+x(\theta)x''(\theta)+y(\theta)y''(\theta)=0,$$
Which means
$$x(\theta)x''(\theta)+y(\theta)y''(\theta) = \text{negative constant}.$$
From $x''=Kx$ and $y''=Ky$, this implies that $K=\text{negative constant}$.  That is, in particular, $y''= Ky$ for some negative constant $K$.
