Finding $\sum_{i=1}^{n-1}\frac1{\cos a_i \cos a_{i+1}}$, where $a_1$, $a_2$, $\ldots$, $a_n$ form an arithmetic progression

Question

Please provide me some hint not solution as I want to do this one with my own hands.

Here is the problem:

The numbers $a_1$, $a_2$, $\ldots$, $a_n$ form an arithmetic progression. Find $$\sum_{i=1}^{n-1}\frac1{\cos a_i \cos a_{i+1}}$$

What I tried :

trigonometry

$$2\cos A\cos B=\cos(\text{sum}/2)\cos(\text{diff}/2)$$ Where $\text{sum}=A+B$, $\text{diff}=A-B$. To simplify the denominator. $$\sec^(A)=1+\tan^2(A)$$ and other trigonometric formulas

Hid and trial: The answer should be $(n-1)\times\text{something}$.
I think we have to add and subtract something to start a chain rxn. May be some manipulations and formula which contains $\sec(\theta)$ terms.

https://math.stackexchange.com/questions/464031/find-the-sum-frac1-cos0-circ-cos1-circ-frac1-cos1-circ-cos2-cir — lab bhattacharjee, Jun 19 '20 at 18:04

score 0 · Answer 1 · edited Jul 09 '22 at 09:33

0

We get the series summation $\sec a_i\sec a_{i+1}$, which very easily simplifies to

$$ \displaystyle\sum\frac{(\tan a_i-\tan a_{i+1})}{\sin(\text{common difference})}$$

But I’d like to see some other approaches too.

edited Jul 09 '22 at 09:33

answered Jun 19 '20 at 16:33

Akash

77

1

This series telescopes. That will be quite helpful. – user317176 Jul 09 '22 at 10:01

Finding $\sum_{i=1}^{n-1}\frac1{\cos a_i \cos a_{i+1}}$, where $a_1$, $a_2$, $\ldots$, $a_n$ form an arithmetic progression

1 Answers1