Talk:Cosine of Sum

The Geometric Proof is the only valid proof.

Proof by Algebra and Proof by Euler's Formula are both invalid proofs.

The reason I say this is because, if you truly respect the notion of rigorous proof, these two proofs I criticize are literally circular in logic.

In order to prove Euler's Formula or the Maclaurin Series for sine and cosine, you must know how to take the derivatives of sine and cosine. There is simply no other way to prove Euler's Formula or the Maclaurin series except en route of the derivatives of sine and cosine.

Furthermore, how do you prove the derivatives of sine and cosine? Via the limit definition of a derivative, one must already know how to find the sine and cosine of a sum and a difference.

Thus, what we have here is a case in which we prove one identity with a second and a second identity with the first.

This proof page uses consequence of truth as a premise to prove. The premises of these proofs are "corollaries", so to speak, if you delve deep enough into math, far beyond the level of math in which these trigonometric identities would be proved.

Don't get me wrong, these so-called "proofs" are marvelous demonstrations... but a demonstration is all that it is worth.

I wonder how many other Proof Wiki pages are logically corrupted?!?!

--CogitoErgoCogitoSum 20:34, 8 December 2009 (UTC)

Reply
That's a bit dogmatic. Have you traced through the chain of proofs? For the "proof by algebra", the sine and cosine are here defined by the series given. They are not derived from Euler's Formula or Maclaurin series. They are defined as they are, and the properties of sine and cosine are deduced directly from those definitions.

As an approach it's as valid as the geometrical approach, but leads itself more easily to an axiomatic approach from which you start with ZFC, define the natural numbers (still work in progress), then integers and so on up to reals, then derive the whole of analysis from understanding basic properties of the real numbers.

From that point of view, the reasoning is not circular.

Alternatively, you can start with the geometrical proofs, as you suggest, but to do this one needs to get the basics of geometry out of the way (and we're barely started on Euclid yet, haven't finished his first book, we're lazy so-and-so's on this site, we get bored easily).

Or you can start with Euler's Identity, which can be derived from a different axiomatic framework (but as I didn't put that proof up, I can't vouch for the details of what direction this will come from).--Matt Westwood 22:40, 8 December 2009 (UTC)

Reply
You completely missed my point and failed to appreciate anything I said. Should I waste my time debating it with you?

Yes, it is dogmatic. But guess what... this is a MATH PROOF. It shouldnt be anything BUT dogmatic.

The so-called proofs that I criticized are indeed invalid. They are circular when you "trace through the chain of proofs", as you said. If you dont believe me then I recommend you trying to prove it yourself, step by step, a linear progression from one proof to the next subsequent proof. You WILL see what I mean.

Just because the proof by algebra section didnt specifically mention the Maclaurin series is irrelevant. Those series' are based on / proved by the Maclaurin series and thus the Maclaurin series' are implied.

Before you pose yet another half-baked rebuttal, try proving it yourself. Step by step, as I instructed.

I bet you that you cannot arrive at those two series for sine and cosine without using the Maclaurin series, thus the Taylor series, thus the derivative of sine and of cosine. And then, when you realize that, try proving the derivative of sine and cosine without first knowing how to split the sine of a sum of angles.

Its absurd to axiomize the rules / equivalences that youre using. Because no one mathematical concept can have multiple simultaneous definitions without that relationship first being proven. There is only one, original definition/conceptualization from which all other theorems derive. This approach is fundamental to any rigorous proof method. How do you rationally go from sine equals opposite divided by hypotenuse to an infinitely alternating series in one fell swoop? Sure, you could theoretically define another second function as that series, but such of a definition would either be arbitrary or based on some other purposeful definition, but its relationship to the sine function would be unproven.

I truly dont see why this is escaping you.

I never suggested proving all of mathematics in each article. You dont need to prove the basics of geometry in this article. But you do need to ensure that the basics of geometry from which this proof is based dont already rely on the theorem being proved. All theorems used in a proof should be traced backward to more fundamental concepts, proven using simpler mathematics. Math Proof 101. How do you not appreciate that?

Youre essentially suggesting grabbing random ideas and arbitrary definitions out of the air, setting them equal "by definition" to some other previously defined function, and seeing what plays out.

If you dont believe in math rigor in proofs then might I suggest changing the name of the site to exclude the word "proof".

--CogitoErgoCogitoSum 20:43, 10 December 2009 (UTC)

There is more than one direction to approach a concept. Yes, if you like, I did define them by just "grabbing random ideas". By defining the sine and cosine as the infinite series that I did, the trick then is to demonstrate that they fulfil the properties held by those arrived at by means of the geometric approach, and then prove that there are no other functions that obey those properties, thereby providing that ever-missing link. After we've done that, we can then prove that these functions can be obtained by Maclaurin expansion from the geometric definition. This approach is the one taken by K. Binmore in his "Mathematical Analysis".

For every direction of travel there is an opposite direction. The one I took happened to start with axiomatics. I am now working on the other direction. --Matt Westwood 21:52, 10 December 2009 (UTC)

I have to agree with Matt on this one. Following the chain back, the proof from algebraic definitions and the proof from Euler's formula start from taking those series as the definitions of sine and cosine. You might not like those definitions, but they're perfectly valid and frequently used.

To quote Wikipedia: "These identities are sometimes taken as the definitions of the sine and cosine function. They are often used as the starting point in a rigorous treatment of trigonometric functions and their applications (e.g., in Fourier series), since the theory of infinite series can be developed from the foundations of the real number system, independent of any geometric considerations. The differentiability and continuity of these functions are then established from the series definitions alone."

Wolfram is similar: "The sine function can be defined analytically by the infinite sum $\sin x = \sum_{n = 1}^\infty\frac{(-1)^{n-1}} {(2n-1)!} x^{2n-1}$." These are just to illustrate the point, not to say that either source is the be all and end all of math definitions.

The definitions may seem fairly arbitrary out of context, but they are perfectly valid. It took me a while to convince myself that it worked out non-circularly when I first looked at it, but it does. If you honestly think that the proofs are circular when based on the series definitions, please do say exactly how. --Cynic (talk) 04:54, 24 December 2009 (UTC)

Site-structural point
Today, a proof showing only one of the identities was added. It seems that we are up for a refactoring to separate this into Sine of Sum and Cosine of Sum, possibly retaining this page as a convenient reference and web search entry point, which points to Sine of Sum and Cosine of Sum for proofs of the respective statements. --Lord_Farin (talk) 11:50, 3 December 2012 (UTC)

Domain
The domain is never specified. If the domain is $\R$, then the first proof which invokes complex numbers is invalid. If the domain is $\C$, then it is next to impossible to identify the real parts, since $\cos a$ might be complex. --kc_kennylau (talk) 11:51, 28 October 2016 (EDT)


 * I accept your point about the complex case, but why is the proof invalid if the domain is real? --prime mover (talk) 14:31, 28 October 2016 (EDT)


 * I take back my point. The domain is $\R$ then. --kc_kennylau (talk) 20:03, 28 October 2016 (EDT)