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:

x

^{2}+ y

^{2}= 1

It struck me recently that the derivative of sine

^{-1}, which is 1/√(1-x

^{2}), looks like it is somehow gotten from the above; specifically, from the equivalent:

y = ±√(1-x

^{2})

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=

*t*for

_{1}*t*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(

_{1}*t*).

_{1}The inverse, D

^{-1}(), gives the

*t*that we arrive at if we travel a given distance. From this, we can calculate the x-location from the original parametric equation

_{1}_{s: }x = ±√(1-

*t*)! So, here we have it:

_{1}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.

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