Timeline for Sums of arctangents

7 events

when toggle format	what		by	license	comment
Dec 28, 2010 at 18:01	comment	added	Michael Hardy		So the full identity is $$ \sin\left( \sum_k \theta_k \right) = \sum_{\text{odd }n\ge 1} (-1)^{(n-1)/2} \sum_{\|A\|=n} \prod_{i\in A} \sin\theta_i \prod_{i\not\in A} \cos\theta_i. $$
Dec 28, 2010 at 17:59	comment	added	Michael Hardy		Typo. Let's try again: $$ \sum_{\text{odd }n \ge 1} (-1)^{(n-1)/2} \cdots\cdots. $$
Dec 28, 2010 at 11:12	comment	added	Did		Michael: nice identity, but the sums are over $n\ge0$ (odd or even) and $\|A\|=2n+1$.
Dec 28, 2010 at 1:51	history	edited	Michael Hardy	CC BY-SA 2.5	minor clarification
Dec 27, 2010 at 21:38	comment	added	Michael Hardy		I suppose I should add a qualification. The identity would be $$ \arctan a - \arctan b = \text{one of the values of }\arctan\frac{a-b}{1+ab}. $$
Dec 27, 2010 at 19:47	comment	added	Michael Hardy		Modern mathematicians seem very much accustomed to power series but not to trigonometric identities. In the 18th century, Euler began with the identity $$ \sin\left(\sum_k \theta_k\right) = \sum_{\text{odd }n \ge 1} (-1)^n \sum_{\|A\|=n} \prod_{i\in A} \sin\theta_i \prod_{i\not\in A} \cos\theta_i $$ and derived the power series for sine from it by saying $$ \theta = \frac{\theta}{n} + \cdots + \frac{\theta}{n} $$ where $n$ is an infinitely large integer, then applying the identity above, then saying that since $n$ is infinitely large, $\sin(\theta/n) = \theta/n$ and $\cos(\theta/n) = 1$.
Dec 27, 2010 at 19:01	history	answered	Michael Hardy	CC BY-SA 2.5