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Math Help - Trig Identities simplification

  1. #1
    Member jacs's Avatar
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    Trig Identities simplification

    I have an identity i need to simplify, i got the left hand half of it done, but i am not at all sure what to do with the right half of it. Have tried numerous things but none of them are working for me.


    thanks for any help
    jacs
    Attached Thumbnails Attached Thumbnails Trig Identities simplification-trig.gif  
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jacs View Post
    I have an identity i need to simplify, i got the left hand half of it done, but i am not at all sure what to do with the right half of it. Have tried numerous things but none of them are working for me.


    thanks for any help
    jacs
    What exactly is the identity you are trying to show?

    -Dan
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  3. #3
    Member jacs's Avatar
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    It just said simplify, that was it.

    So perhaps that is as simple as I should be able to get?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jacs View Post
    I have an identity i need to simplify, i got the left hand half of it done, but i am not at all sure what to do with the right half of it.
    What are the two sides? You only gave us one part of the problem! What was the original problem you were given?

    (An identity is an equation that is satisfied for all possible values of the argument, thus there are two sides to this and not just a simplification problem. Sometimes by seeing both sides of a proposed identity one can guess at what to do.)

    -Dan
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jacs View Post
    I have an identity i need to simplify, i got the left hand half of it done, but i am not at all sure what to do with the right half of it. Have tried numerous things but none of them are working for me.


    thanks for any help
    jacs
    i don't have time to complete this now, but maybe this will give you some ideas.

    \sqrt {1 - \cos 2x} \sqrt {1 + \tan 2x}

    = \sqrt {(1 - \cos 2x)(1 + \tan 2x)}

    = \sqrt {1 + \tan 2x - \cos 2x - \cos 2x \tan 2x}

    = \sqrt {1 + \frac { \sin 2x}{ \cos 2x} - \cos 2x - \cos 2x \frac { \sin 2x}{ \cos 2x}}

    = \sqrt { 1 + \frac { \sin 2x}{ \cos 2x} - \cos 2x - \sin 2x}

    = \sqrt { \frac { \cos 2x + \sin 2x - 1 - \sin 2x \cos 2x}{ \cos 2x}}

    = \sqrt { \frac { \cos 2x - 1 - \sin2x ( \cos 2x - 1)}{ \cos 2x}}

    = \sqrt { \frac {( \cos 2x - 1)( 1 - \sin 2x)}{ \cos 2x}}

    ....and the battle continues. Tune in next time for the next exciting episode of ... *sound effects, dum dum dduuuummm!* TRIG SIMPLIFICATION!


    EDIT: You know, i'm thinking that we may have been able to jump from the second line to the last line...Oh well, i have to leave

    EDIT: Hmm, topsquark believes it was a trig identity proof and not Trig simplification. well, that would be nice, it would give us some direction what to do exactly if we had another side to work towards
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  6. #6
    Member jacs's Avatar
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    There was no other side to the problem. It just said simplify, amd that was the complete problem. (it was in the section of the book that said identities, which is why i called it that)

    I think when i siad left half it might have been misleading, I just mean out of the two square roots, I could deal with the left one without any problem, just wasn't sure what to do with the right one.

    Jhevon, didn't even think to try your approach by combining into a single square root.
    LMAO, love the narrative at the end!!!!
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jacs View Post
    There was no other side to the problem. It just said simplify, amd that was the complete problem. (it was in the section of the book that said identities, which is why i called it that)

    I think when i siad left half it might have been misleading, I just mean out of the two square roots, I could deal with the left one without any problem, just wasn't sure what to do with the right one.

    Jhevon, didn't even think to try your approach by combining into a single square root.
    LMAO, love the narrative at the end!!!!
    yeah, my problem is that what i ended up with doesn't look any simpler than what i started with. but maybe i gave you a few ideas that can help you get a better grip on the problem. i think you should make use of the turning it into one squareroot thing and using sine/cosine for tangent, instead of the complicated double angle formula
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