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Math Help - Trigonometric Identities : Miscellaneous Problem

  1. #1
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    Lightbulb 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
    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
    Last edited by zikcau25; April 18th 2012 at 11:01 AM.
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  2. #2
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    Re: Trigonometric Identities : Miscellaneous Problem

    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}
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  3. #3
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    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).
    Last edited by zikcau25; April 19th 2012 at 09:26 AM.
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  4. #4
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    Thumbs up 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.

    Trigonometric Identities : Miscellaneous Problem-01.png


    *CONVENTIONAL SOLUTION & PROOF:
    Trigonometric Identities : Miscellaneous Problem-02.png


    Now I came up with a solid explanation.

    Hope you appreciate my work. Thank you.
    Last edited by zikcau25; April 19th 2012 at 09:43 AM.
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