In the rant I wrote at


I asked: Are these four identities the first four terms in a sequence that continues?

This referred to the identities in the last bullet point above that question.

While we're at it, is there any intuitive geometric interpretation of the identity involving $f_2$?

OK, here are the functions involved:

$$ \begin{align} f_0(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{even }n \ge 0} (-1)^{n/2} \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\ f_1(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{odd }n \ge 1} (-1)^{(n-1)/2} \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\ f_2(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{even }n \ge 2} (-1)^{(n-2)/2} n \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\ f_3(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{odd }n \ge 3} (-1)^{(n-3)/2} (n-1) \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\ f_4(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{even }n \ge 4} (-1)^{(n-4)/2} n(n-2) \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\ f_5(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{odd }n \ge 5} (-1)^{(n-5)/2} (n-1)(n-3) \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\ f_6(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{even }n \ge 6} (-1)^{(n-6)/2} n(n-2)(n-4) \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\ f_7(\theta_1,\theta_2,\theta_3,\dots) & = \sum_{\text{odd }n \ge 7} (-1)^{(n-7)/2} (n-1)(n-3)(n-5) \sum_{|A| = n} \prod_{i\in A} \sin\theta_i\prod_{i\not\in A}\cos\theta_i \\ & \vdots \end{align} $$ In each function the coefficient kills off the terms involving values of $n$ smaller than the index, so that for example we could have said "$\text{odd }n \ge 1$" instead of $\text{odd }n \ge 7$ and it would be the same thing.

Now some facts:

  • Each $f_k$ is a symmetric function of $\theta_1,\theta_2,\theta_3,\dots$.

  • 0 is an identity element for each of these functions, in the sense that $$f_k(0,\theta_2,\theta_3,\dots) = f_k(\theta_2,\theta_3,\dots).$$

  • $f_k(\theta_1,\theta_2,\theta_3,\dots) - f_k(\theta_1+\theta_2,\theta_3,\dots) = \left. \begin{cases} k & \text{if }k \ge 2\text{ is even} \\ k-1 & \text{if }k \ge 2\text{ is odd} \end{cases} \right\}\cdot \sin\theta_1\sin\theta_2 f_{k-2}(\theta_3,\theta_4,\dots)$ and $=0$ if $k = 0\text{ or }1$.

Now the sequence of identies: $$ \begin{align} f_0 & = \cos(\theta_1 + \theta_2 + \theta_3 + \cdots) \\ f_1 & = \sin(\theta_1 + \theta_2 + \theta_3 + \cdots) \\ \text{If } \sum_{i=1}^\infty \theta_i = \pi,\text{ then } f_2 & = \sum_{i=1}^\infty \sin^2\theta_i=\frac{1}{2} \sum_{i=1}^\infty (\cos(2\theta_i)-1)\\ \text{If } \sum_{i=1}^\infty \theta_i = \pi,\text{ then } f_3 & = \frac{1}{2} \sum_{i=1}^\infty \sin(2\theta_i) \end{align} $$ The QUESTION is whether these are the first four identities in a sequence that continues beyond this point.

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    $\begingroup$ Please try to make your question self-contained by not including a link that the reader must click on in order to parse it. This will greatly increase the readership of your question and thus the likelihood of getting a good answer. $\endgroup$ – Pete L. Clark Jun 14 '10 at 14:46
  • $\begingroup$ Looking at the question and the answer below, I was intrigued how all these "trigonometric identities" could imply "the irrationality of pi". I am disappointed: there is nothing about irrationality nor about pi there... :-( $\endgroup$ – Wadim Zudilin Jun 14 '10 at 15:17
  • $\begingroup$ Actually, if you're looking only at the question and not at the external link, you won't see any of that, and the external link has its own external link to a Wikipedia article that contains Mary Cartwright's proof of the irrationality of $\pi$, so I am somewhat guilty of non-self-containment as suggested above. $\endgroup$ – Michael Hardy Jun 14 '10 at 16:11
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    $\begingroup$ The nLab page is really hard to read. $\endgroup$ – Qiaochu Yuan Jun 14 '10 at 20:30
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    $\begingroup$ Michael, you still have a chance to edit your question by adding the necessary contents and, of course, the trig identity itself. $\endgroup$ – Wadim Zudilin Jun 14 '10 at 23:27

With binomial theorem, the products on the right have closed form $$ \sum_{|A|=n} \prod_{i \in A} \sin \theta_i \prod_{i \notin A} \cos \theta_i = \prod_{i=1}^n ( \sin \theta_i + \cos \theta_i )^n = \prod_{i=1}^n \sqrt{2} \sin (\theta_i + \pi/4)^n$$ So we'll let $x = \sin \theta + \cos \theta$ so the first sum looks like $$ \sum_{n \geq 0, even} (-1)^{n/2} \prod_{i=1}^n x_i = 1 - x_1x_2 + x_1x_2x_3x_4 - \dots$$ This is not symmetric in the x's.

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    $\begingroup$ John, I don't think this is right. What is true is that $$f_0+if_1=\prod_j(\cos\theta_j+i\sin\theta_j)$$ from which the $f_0$ and $f_1$ identities drop out immediately. $\endgroup$ – Robin Chapman Jun 14 '10 at 15:04
  • $\begingroup$ The first two identities are of course universally known. The third one might be entirely novel for all I know. The simplest non-degenerate special case of the fourth one can at least be facetiously referred to as "well known", and possibly is actually well known within certain communities---I don't really know. (For the sake of at least a little bit of "self-containment", the simplest non-degenerate special case of the fourth one says that if $\alpha + \beta + \gamma = \pi$ then $4\sin\alpha\sin\beta\sin\gamma = \sin(2\alpha) + \sin(2\beta) + \sin(2\gamma)$.) $\endgroup$ – Michael Hardy Jun 14 '10 at 16:19
  • $\begingroup$ Yeah I misread the A as running through subsets of {1, 2, \dots, n}. In fact A runs through all n-element subsets of the natural numbers. $\endgroup$ – john mangual Jun 14 '10 at 17:46

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