It has recently come to my attention that the usual unit circle derivation of the elementary trigonometric functions isn't considered rigorous enough. Apparently, this has to do with problems surrounding the definition of arc length.

Could somebody explain, to a freshman mathematics major, why this isn't rigorous enough. With emphasis on freshman: relatively simple explanations would be appreciated over highly abstract ones as I'm still only halfway through my "introduction to analysis" course.

And in contrast - while we're at it - why is deriving the sine function from the integral definition of the arcsine $(*)$

$$\arcsin(x) =\int_0^x \! \frac{1}{\sqrt{1-t^2}} \, \mathrm{d}t \; \; (*)$$

considered to be more rigorous? In order for somebody to know this definition, it would seem to me that you'd have to prove that this is the integral of the arcsine function. And in order to prove this property, you'd have to have a clear definition of an arcsine function that doesn't rely on integrals, which brings you back to the sine function (defining the arcsine as the inverse of a sine function).

So you'd end up using property $(*)$ to create a rigorous definition for trigonometric functions even though this property was derived non-rigorously, thus inviting the question "why wasn't the definition using arc length sufficient if the rigorous approach appears to invoke non-rigorous ideas itself?"?

Again, please keep it within the bounds of what a freshman could understand, or at least try to do so. Much appreciated!

  • 6
    $\begingroup$ If (*) is taken to be the definition of $\arcsin$, then $\sin$ can be defined to be its inverse (which exists, as the integrand is positive). You don't have to prove anything; it's what you're defining $\arcsin$ as. $\endgroup$ – anomaly Jun 7 '16 at 18:53
  • 4
    $\begingroup$ In a nutshell: if you state that $\sin(x)$ is the height of the point on a unit circle corresponding to arclength $x$, then you inevitably have to answer: "what is arclength"? The integral definition allows you to state what you're talking about without answering that question. $\endgroup$ – Omnomnomnom Jun 7 '16 at 18:56
  • $\begingroup$ A simpler analogy for you might be the one where $\log x$ is defined as the integral of the reciprocal function, and $\exp$ is defined as the inverse. This is a perfectly acceptable way of proceeding; the miracle is that this agrees with all the other known ways to define these functions. $\endgroup$ – J. M. is a poor mathematician Jun 7 '16 at 18:57
  • 4
    $\begingroup$ To me, it seems that you have some conflicts between what you have learned in high school and what you are being told now. You should be very careful in using the words when you are studying a pre-learned topic with a different approach. Here you are defining $\arcsin$ with an integral. Take a look at this post. It will be helpful, I think. :) $\endgroup$ – H. R. Jun 7 '16 at 18:58
  • 1
    $\begingroup$ The problem is not the $\sin$, but the notion of angle. How should we relate this intuitive notion from high school geometry to functions defined on an infinite time axis? $\endgroup$ – Christian Blatter Jun 7 '16 at 19:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.