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Does anyone know of any "primitive" proofs of the integrals of sine and cosine, i.e. ones not making use of the fundamental theorem of calculus?

EDIT: Since apparently people believe I have not provided enough "context" here is an attempt of re-stating the question. Does anyone know of any first principles proof of the fact that: $$ \int_a^x\cos(t) \,dt = \sin(x) + C $$ without relying on the FTC or derivates? Ideally this proof would use a suitably simple definition of $\sin$ and $\cos$, for instance the unit circle definition.

Peatherfed
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    That "primitive" proof relies on how you define formally the sine and cosine functions. – azif00 Jun 01 '21 at 02:13
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    The fact that $\sin$ is an antiderivative of $\cos$ does not use the fundamental theorem of calculus. The fundamental theorem of calculus tells us that for all $a\in\mathbb{R}$$$\frac{d}{dx}\int_{a}^{x}\cos t , dt = \cos x , ,$$and also that for all $a,b\in\mathbb{R}$$$\int_{a}^{b}\cos x , dx = \sin(b)-\sin(a) , .$$ – Joe Jun 01 '21 at 08:46
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    $$h\sum_{k=0}^{n-1}e^{ikh}=\dfrac xn\dfrac{e^{inh}-1}{e^{ih}-1}\to\frac{e^{ix}-1}i$$ –  Jun 01 '21 at 16:14
  • @azif00 That is true, and I should have made clear in my post which one I was using. Considering I'm asking for a primitive proof of the result, I thought it was clear that using a rather simple (i.e. unit circle) definition of sine and cosine was what I was doing. – Peatherfed Jun 01 '21 at 17:57
  • @Joe Every single proof I've seen uses the fundamental theorem of calculus, either directly or indirectly. How else do you propose to use the fact that d/dx sin(x) = cos(x) means that the integral of cos(x) is sin(x)? – Peatherfed Jun 01 '21 at 18:00
  • @Peatherfed: If you are talking about indefinite integrals, then by definition$$\int \cos x , dx$$refers to the set of functions $F$ such that $F'=\cos$. Since $\sin'=\cos$, and the set of antiderivatives differ by a constant, we write$$\int \cos x , dx = \sin x + C , ,$$(Although a more pedantic notation would be$$\int \cos x , dx = {F:F(x)=\sin x + C \text{ for all $x$}, C\in\mathbb{R}} , .)$$The fundamental theorem of calculus tells us the link between antiderivatives (or indefinite integrals) and definite integrals. I don't see any reason to invoke it here. – Joe Jun 01 '21 at 18:37
  • @Joe $\int \cos x , dx = {F : F(x)=\sin x + C \text{ for } \color{red}{\rm some} \text{ $C \in\mathbb{R}$ and all $x \in\mathbb{R}$}}$. – azif00 Jun 01 '21 at 18:53
  • @Joe Except that definition is informed by the FTC, otherwise, it would be a completely useless definition. I agree that the anti-derivative of e.g. cos(x) is sin(x), but I am not asking for the anti-derivative in my question, I'm asking for the integral. And before you protest that they are in this case the same, since I am implicitly asking for an indefinite integral, I reiterate that they are the same by definition, since the definition is informed by the FTC. Perhaps I was being unclear in my question, but I think this was readily understood. – Peatherfed Jun 01 '21 at 19:03
  • @Peatherfed: I agree that the power of antiderivatives is only understood when the Fundamental Theorem of Calculus has been proven. So, to be clear, you are asking for a 'first principles' way of working out$$\int_{a}^{x}\cos t , dt , ? \tag{}\label{}$$And this method shouldn't rely on the fact that $\eqref{*}$ is an antiderivative of $\cos$. Is that correct? – Joe Jun 01 '21 at 19:34
  • @Joe Yes, that is correct. – Peatherfed Jun 01 '21 at 19:43
  • @Peatherfed: In the body of your question, you might want to change the variable inside the integrand to '$t$' or something similar. – Joe Jun 01 '21 at 22:28
  • @azif00: Are you saying that I should have put a quantifier over $C$ because I am introducing this variable within the set builder notation itself? – Joe Jun 01 '21 at 22:44
  • @Joe Yeah I blanked on that one, the question is now edited with a proper variable inside the integral. Regardless, for some reason, the question was deemed to "not have enough context". I'm not sure how much more context I could possibly provide. It was a straightforward question. I feel like a lot of the people voting to close it just didn't know of any such proof, and as such the question must be bad. – Peatherfed Jun 01 '21 at 23:14
  • Yes, "not have enough context" is what is the problem. First, you did not define what you mean by integral. Only when I asked for one you mentioned upper and lower Riemann sums. Also you are still not completely precise on exactly how you define sine and cosine. Usually these issues do not arise because all definitions should be essentially equivalent, but when you want a "primitive" proof, you have to be precise about what is allowed to be assumed. Also, the question is not that great to begin with. – Somos Jun 01 '21 at 23:43
  • @Somos The question has been edited so that my question should be more clear. The only way to provide more context would be to list literally every definition, and that is ridiculous. It's a soft question. The way I define integral is the standard way, so I'm not sure why you think it was necessary for me to explicitly mention upper and lower Riemann sums. Second, I am being precise in how I define sine and cosine while leaving options open for other definitions as long as they are in the spirit of the question. – Peatherfed Jun 01 '21 at 23:52
  • The Wikipedia article Riemann integral mentions several variations of definition of an integral. You need to be precise about which variations you want to allow to be used. You can refer to standard sources of definition such as Wikipedia. The actual definition of sine and cosine is not important in this context. I think it is not important here which definition is used. So the question is how to prove $\int_0^x \cos(t)dt = \sin(x)$ with a suitable variation of integral. – Somos Jun 02 '21 at 00:05

1 Answers1

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One approach is to directly use Riemann sums. Using the result from MSE question 17966 ("How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?") let $$ S_n := \sum_{k=0}^{n-1} \sin\left(a+\frac{(b-a)k}n\right)\frac{b-a}n \\ = \sin\left(\frac{b-a}2\right)\sin\left(\frac{b+a}2-\frac{b-a}{2n}\right) \frac{\frac{b-a}n}{\sin(\frac{b-a}{2n})}$$ be the left Riemann sum over $\,n\,$ subintervals of $\,[a,b].\,$ Given that $\,\sin\,$ is continuous and that $\,\lim_{x\to 0} \frac{\sin x}x = 1,\,$ check that $$ \lim_{n\to\infty} S_n = 2\sin\left(\frac{b-a}2\right)\sin\left(\frac{b+a}2\right). $$ This agrees with the value of $$ \int_a^b \sin(x)dx = \cos(a)-\cos(b) $$ using a standard trigonometric identity for difference of two cosines.

Somos
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  • This is a good start. Ideally, I'd like to drop reliance on limits altogether, but perhaps this is something I can work out in my own time. – Peatherfed Jun 01 '21 at 18:00
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    @Peatherfed How are you going to define integrals without limits? It is possible but you have to tell us how you define integrals. – Somos Jun 01 '21 at 22:29
  • Like anyone else defines integrals without limits. Take the set of lower Riemann sums, and take its supremum, and then take the set of upper Riemann sums and take its infimum. If these are equal, then that real number is the integral. I prefer this over involving limits, since if you do limits you have to involve annoyances such as mesh size of your partitions. – Peatherfed Jun 01 '21 at 22:42
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    @Peatherfed Thanks for that clarification. However, proving anything about infimum and supremum involves epsilon arguments which is morally equivalent to taking limits. – Somos Jun 01 '21 at 22:51
  • No, you can prove things about infimum and supremum without epsilon arguments, and without using limits. – Peatherfed Jun 01 '21 at 22:56