[Math] Trigonometric identities

special functionstrigonometric-sums

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.

Best Answer

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.

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