If we define $\\sin x$ as series, how can we obtain the geometric meaning of $\\sin x$?

Question

In Terry Tao's textbook *Analysis*, he defines $\\sin x$ as below: > 1. Define rational numbers > 2. Define Cauchy sequences of rational numbers, and equivalence of Cauchy sequences > 3. Define reals as the space of Cauchy sequences of rationals modulo equivalence > 4. Define limits (and other basic operations) in the reals > 5. Cover a lot of foundational material including: complex numbers, power series, differentiation, and the complex exponential > 6. Eventually (Chapter 15!) define the trigonometric functions via the complex exponential. Then show the equivalence to other definitions. **My question** is how can we obtain the geometry interpretation of $\\sin x$, that is, the ratio of opposite side and hypotenuse.

Answer 1

Easier, less rigorous justification: You probably have seen this justification in a calculus II course where you discussed power series representations of infinitely differentiable functions (like sine and cosine). Any calculus textbook will show you--probably by Taylor's Remainder Theorem (or something closely related)--that the power series representations for sine and cosine converge to sine and cosine, respectively. How does this connect back with your sine and cosine functions of trigonometry? Well in your calculus books the derivation of the derivatives of sine and cosine are found by appealing to these geometric definitions, where the core limit \lim_{x \to 0} \sin (x)/x =1 is established via an argument using areas of geometric figures and those very "trig" definitions of sine and cosine. In your calculus sequence (for scientists and engineers) you are just "given" the trig functions, in other words, they aren't defined in the rigorous way as the book by Tao does. So in other words you have seen an explanation for why those series are indeed actual representations for your geometrically defined sine and cosine functions, it mostly likely wasn't explicated well. I, too, took a long time to figure out. But asking this question shows you are really asking the right questions to deeply understand the subject.

More sophisticated, difficult, but more rigorous answer:

CAUTION: The answer to this question is more technical than most calculus students could follow. But you (the original poster) are at the level of "real analysis" and reading from Terry Tao's book, so you should have the prerequisite knowledge to follow.

The question you have asked is a very good question that most students overlook. Students don't often reconcile higher levels of mathematical sophistication with the simpler ideas they learned during an earlier encounter.

Seeing that you are reading an analysis textbook by Terry Tao, you are familiar with elementary differential equations and the language of calculus. The explanation I give is fairly rigorous, but you will see there is no way we could present these ideas to a trig student, at least not without teaching them calculus, ha.

Notice that when we define the trigonometric functions in a trig class it is usually via a "unit circle" approach. Now by modern standards these aren't rigorous definitions in the sense that they aren't built up from functions defined in the guise of set theory, which is what Tao is trying to do in his excellent book. For trig students, once you "believe" in the arc length of a sector of a circle (again for students not understanding calculus, arc length isn't rigorously defined) you know there should be an association between the x or y coordinates of points on the circle and the arc length s. This is how the "unit circle" approach goes about "defining" the trig functions. Remember that in the unit circle approach the angle θ is just the arc length s, so I use θ in place of s.

It's actually easier to discuss these ideas with the cosine function, rather than the sine function, so I'll discuss how this is done with the cosine function.

Unfortunately, this comment section doesn't support LaTeX markup so the written math isn't going to be aesthetically pleasing, sorry.

First, define the "theta function" θ(x) = \int_x^1 dt / \sqrt{1-t^2)}. Where do I get this integrand? Just look up your old formula for the arclength along a curve y=f(x), where y=f(x) in our case is just y = \sqrt{1-x^2} since we are finding the arclength along the unit circle from x to 1 (again the arclength s and our angle θ are the same for us).

Now what does θ(x) do? It defines the crucial relationship between the the arclength and the x-coordinate, which is exactly what your old cosine function did in that trig class. In other words, x = cos(θ) (familiar trig cosine function) or θ(x) = arccos(x).

Now, I'm sure that in that analysis class you analytically derived the basic identities: sin^2(x) + cos^2(x) = 1 and the derivative of cos(x) is -sin(x). Now with y = cos(x) (analytic cosine function), we have a solution to the differential equation y' = - \sqrt{1-y^2} (technically you need to know sin(x) is nonnegative on some interval [0,a) ).

Now using the fundamental theorem of calculus (differentiating integrals), we have θ'(x) = - 1/ \sqrt{1-x^2}.

Now consider the function θ(y(x)). By the chain rule d/dx {θ(y(x))} = θ'(y(x)) • y '(x) = 1. So, we must have (by the mean value theorem of derivatives if you wish) that θ(y(x)) = x + c for some constant c. Now since y(0) = cos(0) = 1 (following from the power series expansion of cosine) we have, 0 = θ(1) = c, i.e. our constant c is actually just 0.

So we have θ(y(x)) = x, meaning y(x) = θ^{-1}(x) (inverse function of θ). And this last equation is what we wanted to show. y(x) is your power series representation of the cosine function (defined analytically by Tao) while the inverse of our theta function θ(x) (which is our geometric arccosine function defined via integral) is just the geometric cosine function. Voila, the two function are indeed the same. Now any geometric intuition you have with the cosine function can be transferred over to whenever you deal with the analytically defined cosine function.