Monday, February 15, 2010

Defining Sine

I don't know if this is a sighn of things to come, but the first post is not entirely on-topic. :)
 I suppose I should link back to my older blog before I begin:

http://dragonlogic-ai.blogspot.com/

 The move roughly indicates me realizing that I now know too much logic & math to be able to write to a general audience without a fair amount of exposition, and also enough that I want to correct much of what my previous self said.
 Anyway.
 It's bothered me for some time that the functions sine and cosine are given, in a sense, without definition (in textbooks, that is). We didn't even get a method for calculating these, as with the algebraic functions we'd learned.  I know, of course, that they are defined in terms of angles and the unit circle. Angles, however, are not given an algebraic definition either... fortunately, the unit circle is defined:

x2 + y2 = 1

It struck me recently that the derivative of sine-1, which is 1/√(1-x2), looks like it is somehow gotten from the above; specifically, from the equivalent:

y = ±√(1-x2)

Rather than look for a proof of this derivative, I decided to see if I could figure it out for myself.
 Now, if I had been given the task of finding the derivative of sine-1, I would have taken the implicit derivative, which gives you the derivative of any inverse function in terms of the derivative of the original function:

f(f-1(x)) = x
f(f-1(x))' = x'
f'(f-1(x))f-1'(x) = 1
f-1'(x) = 1/ f'(f-1(x))

 So, for any f, you can find the inverse's derivative if you know the inverse and the derivative.
 In particular,

sine-1'(x) = 1/cos(sin-1(x))

Of course, this is not nearly as useful as the derivative given in my textbook (for finding integrals of algebraic functions, that is).
 So, the question I ask myself is: How can we define sine-1 algebraically?
 Intuition: The "angle" is a measurement of the length along the unit circle that must be traveled, starting at (1,0) and going counter-clockwise,  to get to a particular x-location.
 From the 2nd form for the equation for the unit circle, it's fairly simple to get a parametric equation system:

y = ±√t
x = ±√(1-t)

In order to keep the distance we're moving steady as we increase t, we want to stick the derivatives of x and y in t into the distance formula:

 √(¼t-1 + ¼(1-t)-1)

Now, the integral of this from t=0 to t=t1 for t1 between 0 and 1 is what we're interested in. This is a measure of the distance we've traveled on the unit circle as t has changed. Denote this integral as a function, D(t1).
 The inverse, D-1(), gives the t1 that we arrive at if we travel a given distance. From this, we can calculate the x-location from the original parametric equations: x = ±√(1-t1)! So, here we have it:

sin-1(x) = ±√(1-D-1(x))

We can then proceed to define sin as the inverse of sin-1.

If anyone can find any mistakes in this, thanks in advance. Rightly speaking, to get an actual purely algebraic definition, I need to go and find that integral. Perhaps another day. :)

Edit-- Fixed some errors thanks to Daniel Demski.