Trigonometric Identities : Miscellaneous Problem

**zikcau25** · April 18th 2012, 11:40 AM

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

Trigonometric Identities : Miscellaneous Problem-01.png

2. Working ...

Trigonometric Identities : Miscellaneous Problem-02.png

3. Simplifying ...

Trigonometric Identities : Miscellaneous Problem-03.png

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.

Trigonometric Identities : Miscellaneous Problem-x.png

Trigonometric Identities : Miscellaneous Problem-04.png

**Soroban** · April 18th 2012, 06:58 PM

Hello, zikcau25!

Welcome aboard!

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

I would start with the right side . . .

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

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

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

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

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

We can start with the left side, too . . .

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

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

**zikcau25** · April 19th 2012, 08:49 AM

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).

**zikcau25** · April 19th 2012, 10:07 AM

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.

*CONVENTIONAL SOLUTION & PROOF:

Now I came up with a solid explanation.

Hope you appreciate my work. Thank you.

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