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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?

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The rate at which the height of the tip of the pole increases is a constant times the cosine. –  André Nicolas May 22 '12 at 0:37
    
What do you mean "without using...the unit circle"? The tip of the pole traces out part of a circle, and that fact must be used somehow. –  Jonas Meyer May 22 '12 at 2:30
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1 Answer 1

up vote 3 down vote accepted

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 $K$.

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Very slick, I like it. However, the x^2+y^2 = constant part is dependent upon the Pythagorean theorem, and I don't know of any proof that doesn't (at least implicitly) use sine. Is there one? Is there a way to prove that the motion is harmonic even when it's not a right triangle (where, in the example above, you'd explicitly have to use the cosine rule)? –  Nate May 22 '12 at 4:13
    
@Nate: Thank you. Most proofs of the Pythagorean theorem do not use sine. E.g., look at these. You say "implicitly"--well, you could throw sine into them, but no a priori knowledge of sine is used. What do you mean "even when it's not a right triangle"? What modification of the problem do you have in mind exactly? What do you mean by "you'd explicitly have to use the cosine rule"? There's no cosine rule needed in the above. –  Jonas Meyer May 22 '12 at 4:16
    
I'm just wondering what happens when you generalize to the case where instead of holding the length of the pole (and the 90* angle with the ground) constant, you hold the length of the opposite angle to the ground constant. It seems that, with this method, all triangles (varying one angle and holding another constant) trace out an ellipse, but I'm not sure how to go about proving or disproving such an idea. –  Nate May 22 '12 at 6:52
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