So I know this is a dumb question maybe, but it has been in my head recently. I once told someone, on the topic of introductory calculus, that using l'Hopital's rule to calculate $$\lim\limits_{x\rightarrow 0} \frac{\sin(x)}{x}$$ is circular reasoning, because the limit we are looking for is the definition of the derivative of $\sin(x)$ at $x=0$. They argued that it is not because you can work out the derivative of cosine via its series representation without ever recurring to the definition.

Now, there are two ways that I know of to obtain the series representation of cosine: the obvious one is Taylor expansion, which obviously requires knowledge of all the derivatives, the other one is to prove the following statement:

If there exists two functions $f$ and $g$ such that $f'=g$ and $g'=-f$, then they are unique.

Then you have to kind of get out of the hat the series representations and show that they respect the conditions listed above, which only involves deriving polynomials. Nonetheless, it seems to me that you still have to know the derivatives of sin and cos to then identify the two trigonometric functions with the series representations using their uniqueness.

My question is: is there really a way to obtain the series representation for sin and cos without knowing the derivatives, thus getting rid of the circularity of the reasoning?

This question is by no means practical and it's just a curiosity of mine, I'm not interested in other ways to calculate the limit, which is trivial.


I don't think this is a duplicate, as I said in my comment if you read the question that is linked, you will notice that what the OP means by "circular reasoning" is something on proving the limit using formulas about the area of circles, which has nothing to do with my question about series representations.

  • $\begingroup$ This limit is not derivative of $cos(x)$ at $x=0$, but $cos(0)$ itself. $\endgroup$
    – Przemek
    Jun 25 '17 at 17:49
  • $\begingroup$ @Przemek how embarrassing, I meant of $\sin(x)$, which is what you have to derive to apply the rule, editing now $\endgroup$
    – user438666
    Jun 25 '17 at 17:50
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    $\begingroup$ Chris, if you have a copy of Johnsonbaugh and Pfaffenberger (it's available in Dover) then you'll find a series definition of sine and cosine. In fact, all the usual features of sine and cosine can be derived from such analysis completely devoid of explicit geometric reasoning (as fills the alleged "duplicate"). See page 274-275 or so. Incidentally, I'm currently working on a paper which generalizes such observations to arbitrary associative algebras. So, divorcing sine and cosine from right triangle trig is dear to my heart. $\endgroup$ Jun 25 '17 at 18:23
  • $\begingroup$ @JamesS.Cook thank you for your comments, I tried opening a meta post on this issue, see here, math.meta.stackexchange.com/questions/26553/…. On your second comment, I don't have a copy of that book, I will look for this reference. I had not even thought about the fact that sin and cos are only defined in a geometrical way as far as I know! Can one take the statement I quoted as a definition? If one takes that as the definition, then sin and cos are merely two names for the series, and the circularity is gone, or is it not? $\endgroup$
    – user438666
    Jun 25 '17 at 18:30
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    $\begingroup$ Yes, the circularity is removed. Sine has derivative cosine by the term-by-term differentiation of series theorem. Hence, the usually circular Lhop argument is legit. Of course, for most users of calculus, the option to define sine and cosine via series comes a bit after they see Lhop rule. You can define $\pi$ in terms of series as well. $\endgroup$ Jun 25 '17 at 18:35

We make the following definitions: $$ \sin(x) = x - \frac{1}{3!}x^3+ \frac{1}{5!}x^5+ \cdots = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{2k+1} $$ and $$ \cos(x) = 1 - \frac{1}{2!}x^2+ \frac{1}{4!}x^4+ \cdots = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} x^{2k} + \cdots$$ The series above converge for each $x \in \mathbb{R}$ by the ratio test. A well-known theorem of power series asserts that the derivative of a power series is indeed the power series of derivatives. This is not a trivial theorem, it requires some effort to prove. But, with the term-by-term differentiation theorem is obvious that \begin{align} \frac{d}{dx} \sin(x) &= 1+ \frac{2}{3!}x^2+ \cdots + \frac{2k+1}{(2k+1)!}x^{2k} + \cdots \\ &= 1+ \frac{1}{2!}x^2+ \cdots + \frac{1}{(2k)!}x^{2k} + \cdots \\ &= \cos(x) \end{align} Thus, applying L-Hopital's Rule: $$ \lim_{x \rightarrow 0} \frac{ \sin x}{x} = \lim_{ x \rightarrow 0} \frac{\cos(x)}{1} = \frac{\cos(0)}{1} = 1.$$ Notice, $\cos(0)=1$ and $\sin(0)=0$ are nearly manifest with the definitions I offer at the start of this answer.

Here's an interesting follow-up: how do we prove other aspects of the function theory of sine and cosine via series. For example, how to prove the $2\pi$-periodicity, or adding-angles formulas ? Much is known. I'll leave it at that for now.

  • $\begingroup$ incidentally, I usually say it's circular for students because I don't think they use the definitions offered here. You can also define sine and cosine as the unique solutions $y_1,y_2$ of $y''+y=0$ for which $y_1(0)=1$ and $y_1'(0)=0$ or $y_2(0)=0$ and $y_2'(0)=1$. Of course, $y_1$ is the cosine whereas $y_2$ is sine. $\endgroup$ Jun 25 '17 at 19:07
  • $\begingroup$ Thank you. Just another curiosity, without exploiting the uniqueness of the solutions to the differential equations you mentioned, and the geometric representations, how can one be sure that the series actually are equivalent to the geometric definition? $\endgroup$
    – user438666
    Jun 25 '17 at 19:11
  • $\begingroup$ Well, that is an interesting question. Perhaps you should ask it. I'm not sure I've thought through that in its entirety. Certainly, the fact that $\sin^2 \theta+ \cos^2 \theta =1$ is a big part. So... how to define geometrically though? What does that entail? I'd start with right-triangle trigonometry then extend to other quadrants via sign considerations. Is reproducing polar coordinates enough? Is there more to the "geometry" of sine and cosine? I'm not sure. $\endgroup$ Jun 25 '17 at 19:15
  • $\begingroup$ Related...this page includes (after other proofs) a geometric proof that sin(x)/x limits to 1: proofwiki.org/wiki/Limit_of_Sine_of_X_over_X $\endgroup$
    – Bill Cook
    Jun 25 '17 at 20:12
  • $\begingroup$ Equipped with that fact, you can get the derivative of sine is cosine. $\endgroup$
    – Bill Cook
    Jun 25 '17 at 20:13

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