Trigonometric Identities : Miscellaneous Problem

• Apr 18th 2012, 10:40 AM
zikcau25
Trigonometric Identities : Miscellaneous Problem
It's my first day on this forum.
I was working on a trigonometric identities question this evening, when I surprisingly got stuck.
It was only after that I discover a new simple unlearned principle in the question that I was able to finish it up.

Trigonometric identities most often need binomial expansion approach to it.
In my case, it requires a bit more - Sometimes unexpected challenge show up in the question - and it is exactly what I am going to share with you.
To some this may be a mild thing, but to me today it a great day today. lol

Firstly, I would like to discover the ways proposed by others to deal with the problem,
Before I may post my own solution I discover.

1. Question
Attachment 23638

2. Working ...
Attachment 23639

3. Simplifying ...
Attachment 23640

4. This was where I got stuck,
This is an exemplary case, where I can't find any known/taught (in textbook) clue to solve this problem, but with careful study I could find it.
... I would wish to see your solution to prove this "simple" Algebraic Identity. (Rock)

Attachment 23642

Attachment 23641
• Apr 18th 2012, 05:58 PM
Soroban
Re: Trigonometric Identities : Miscellaneous Problem
Hello, zikcau25!

Welcome aboard!

Quote:

$(\csc\theta - \cot\theta)^2 \;=\;\frac{1-\cos\theta}{1+\cos\theta}$

Multiply by $\frac{1-\cos\theta}{1-\cos\theta}:$

. . $\frac{1-\cos\theta}{1+\cos\theta}\cdot\frac{1-\cos\theta}{1-\cos\theta} \;\;=\;\;\frac{(1-\cos\theta)^2}{1-\cos^2\!\theta} \;\;=\;\;\frac{1-\2\cos\theta + \cos^2\!\theta}{\sin^2\!\theta}$

. . . . . . $=\;\;\frac{1}{\sin^2\!\theta} - \frac{2\cos\theta}{\sin^2\!\theta} + \frac{\cos^2\!\theta}{\sin^2\!\theta} \;\;=\;\;\left(\frac{1}{\sin\theta}\right)^2 - 2\left(\frac{1}{\sin\theta}\right)\left(\frac{\cos \theta}{\sin\theta}\right) + \left(\frac{\cos\theta}{\sin\theta}\right)^2$

. . . . . . $=\;\;\csc^2\!\theta - 2\csc\theta\cot\theta + \cot^2\!\theta \;\;=\;\;(\csc\theta - \cot\theta)^2$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$(\csc\theta - \cot\theta)^2 \;=\;\left(\frac{1}{\sin\theta} - \frac{\cos\theta}{\sin\theta}\right)^2 \;=\;\left(\frac{1-\cos\theta}{\sin\theta}\right)^2 \;=\;\frac{(1-\cos\theta)^2}{\sin^2\!\theta}$

. . . . . . $=\; \frac{(1-\cos\theta)^2}{1-\cos^2\!\theta} \;=\;\frac{(1-\cos\theta)(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)} \;=\; \frac{1-\cos\theta}{1+\cos\theta}$
• Apr 19th 2012, 07:49 AM
zikcau25
Re: Trigonometric Identities : Miscellaneous Problem
You a Genius having dealt with it from beginning to end perfectly ... encountering no obstacle along the way through. Thank you very much.

But in my lengthy approach I confronted a second Puzzle reason for which I make this thread and which I previously simplify, to be analysed, in term of x (step 4. expressed as Algebraic Identity).
• Apr 19th 2012, 09:07 AM
zikcau25
Re: Trigonometric Identities : Miscellaneous Problem - SOLVED
1. Here is my definite solution.

*UNCONVENTIONAL APPROACH:
- Most people would suggest to multiply each of the Binomials on the numerator (LHS) without a CLEAR JUSTIFICATION.
What matters is that it serves to resolve the problem.
What they don't know is that it is unconventional to general algebraic principles and lead to confusion.

Attachment 23654

*CONVENTIONAL SOLUTION & PROOF:
Attachment 23657

Now I came up with a solid explanation.

Hope you appreciate my work. Thank you.