Trigonometric identities

Question

In the rant I wrote at

http://ncatlab.org/nlab/show/trigonometric+identities+and+the+irrationality+of+pi

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 $f_k(0,\theta_2,\theta_3,\dots) = f_k(\theta_2,\theta_3,\dots).$

• Let $\lfloor k\rfloor_\text{even}= 2\lfloor k/2\rfloor$ be the "even floor" of $k,$ i.e. the largest even integer not exceeding $k.$ Then \begin{align} & f_k(\theta_1,\theta_2,\theta_3,\dots) - f_k(\theta_1+\theta_2,\theta_3,\dots) \\[8pt] = {} & \begin{cases} \lfloor k \rfloor_\text{even} \cdot \sin\theta_1\sin\theta_2 f_{k-2}(\theta_3,\theta_4,\dots) & \text{if } k\ge2, \\[6pt] \quad 0 & \text{if } k = 0\text{ or } 1. \end{cases} \end{align}

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

Answer 1

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 $$ \sum_{n \geq 0,\text{ even}} (-1)^{n/2} \prod_{i=1}^n x_i = 1 - x_1x_2 + x_1x_2x_3x_4 - \dots$$ This is not symmetric in the x's.