(This question will at first appear too broad. However, the overall philosophy will be explained below in a way that asks specific questions, which I hope will be conducive to this being a reasonable question.)

I have often wondered to what extent the trigonometrical functions' properties can be proven using only real results (they're real functions, so why would we need complex numbers. The precise formulation of my initial question is

Are there any real trigonometrical identities that require complex-number results to prove?

So, for example, the double-angle formulae can be proven using geometry or matrices, and I think most people would prefer this to the way using complex exponentials, simply because you can do it before introducing complex numbers.

On the other hand, de Moivre's theorem, $$ (\cos{\theta}+i\sin{\theta})^{\alpha} = \cos{\alpha\theta}+i\sin{\alpha\theta}, $$ is excluded because, obviously, it has an $i$ in it. Similarly Euler's formula, and hence the complex expressions of cosine and sine.

At first, I thought that the sum $$ 1+2\sum_{n=0}^{\infty} r^n \cos{n\theta} = \frac{1-r^2}{1-2r\cos{\theta}+r^2} \quad (|r|<1) $$ could only be done with complex numbers, but (after mucking about with the complex exponentials) one finds the quantity $$ C_m =\frac{2r^{m+1}(r\cos{m\theta}-\cos{(m+1)\theta})}{1-2r\cos{\theta}+r^2}, $$ which satisfies $$ C_m-C_{m-1} = 2r^m \cos{m\theta}, $$ and therefore the partial sums are given by $$ 1+2\sum_{n=0}^{m} r^n \cos{n\theta} = 1+C_m-C_0, $$ and $$ 1-C_0 = 1-\frac{2r(r-\cos{\theta})}{1-2r\cos{\theta}+r^2} = \frac{1-r^2}{1+2r\cos{\theta}+r^2}. $$ Since $C_m \to 0$ as $r \to \infty$, we find the answer we were hoping for. Similarly with the product $$ \prod_{k=1}^{n-1} \sin{\left(\frac{k\pi}{n}\right)} = \frac{n}{2^{n-1}}, $$ which can be proved using ingenious use of the factor theorem. (This proof I found on a rather obscure forum somewhere in the depths of the Internet. I'll reproduce it if people are interested)

So, the questions, as I see them, are

  1. Is there a counterexample to the claim that any real trigonometric identity can be proven using only real number techniques?

  2. Is there a general way of proving that no counterexample exists (obviously a long shot, but you never know if there's something like an obscure Galois theory calculation out there...)

  3. If either of those is too general for you, can anyone prove that $$ \prod_{k=1}^{n-1} (1-2r\cos{\left( \frac{k\pi}{n} \right)}+r^2) = \frac{1-r^{2n}}{1-r^2} $$ without using complex numbers (the usual proof uses roots of unity, for fairly obvious reasons).

(If you're wondering where these product formulae come from, G.H. Hardy, for reasons best known to himself, considered the integrals $$ \int_0^{\pi} \log{\sin{\theta}} \, d\theta \qquad \int_0^{\pi} \log{(1-2r\cos{\theta}+r^2)} \, d\theta $$ as among those that should be evaluated by first principles, which leads fairly directly to considering the above products.)

(You can obviously have either the power series, geometrical, or some differential equation definition of the trigonometrical functions, just no quantities that square to $-1$!)

  • $\begingroup$ Your last product is indexed over $k$, but it doesn't contain it anywhere. $\endgroup$ – Akiva Weinberger Apr 23 '15 at 1:51
  • $\begingroup$ Also, it shouldn't be too hard to find $\sum_{n=0}^mr^n\cos n\theta$ using real methods, anyway. $\endgroup$ – Akiva Weinberger Apr 23 '15 at 1:53
  • $\begingroup$ This question is garbage. To begin with, the question itself is trivial. The answer is of course yes. Every proof with complex numbers can be translated into an argument about real numbers. The other problem is that it emanates from the assumption that eliminating complex numbers from an argument is somehow didactically desirable. Quite the opposite. Most of the time complex numbers [planar arithmetic] clarify trigonometric arguments. A historical handicap is the only reason complex numbers are (with the exception of some lucky schools) introduced after trigonometry. $\endgroup$ – Alamos Apr 23 '15 at 2:06
  • $\begingroup$ @Alamos Garbage? Not at all. If this were as easy as you say, real and complex proofs would look near-identical. "Every proof with complex numbers can be translated into an argument about real numbers." What do you mean? Do you have a proof of this statement? And nowhere in my question have I mentioned pedagogy, so I do not see why that is relevant: this is purely a question of proof technique. There are a number of areas of mathematics where not using the complex numbers is desirable. Keep your comments more civil in future: such rudeness is quite unnecessary and detrimental to discussion. $\endgroup$ – Chappers Apr 25 '15 at 17:21
  • $\begingroup$ Every theorem that a mathematician shows to other mathematicians will get criticised if there are any faults with it's logic. At times the criticism will be quite ruthless. So think of it as a complement. You were treated as a real mathematician. $\endgroup$ – steven gregory Sep 26 '15 at 3:31

For every $(a,b), (c,d) \in \mathbb R^2$ define

$$(a,b)+(c,d)=(a+c,b+d) \text{ and } (a,b)\times(c,d) = (ac-bd,ad+bc)$$

It isn't hard to prove that the structure $[\mathbb R^2, +,\times]$ is a field. Nor is it a big suprise that the map $\varphi : \mathbb R^2 \to \mathbb C$ defined by $\varphi(x,y) = x + \imath y$ is an isomorphism. Hence every theorem of complex numbers corresponds to a theorem in the plane $\mathbb R^2$. Hence every theorem using complex numbers can be converted into a theorem that doesn't involve complex numbers.


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