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Math Help - Trig novice needs help; problem ---> a.tan(x) - (b/cos(x)^2) - c = 0

  1. #1
    Newbie SEZHUR's Avatar
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    Trig novice needs help; problem ---> a.tan(x) - (b/cos(x)^2) - c = 0

    This problem bubbled and oozed up from my equations while doing my physics assignment. I'm not even sure this equation is one that we're supposed to answer, seems kinda hard for a non-math intensive paper. My knowledge of the trig identities is completely self taught though, so this may be easier than I imagine.
    Thanks.
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  2. #2
    Master Of Puppets
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    Does a.tan mean arctan?

    Is it cos(x^2) or (cos(x))^2?
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  3. #3
    Newbie SEZHUR's Avatar
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    'a.tan' means 'a multiplied by tan', and its (cos(x))^2
    Sorry for the confusion, not sure what the standard format is on this site, first time user.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by SEZHUR View Post
    'a.tan' means 'a multiplied by tan', and its (cos(x))^2
    Sorry for the confusion, not sure what the standard format is on this site, first time user.
    The standard is that function arguments go in brackets, and brackets go around anything more complicated than a single symbol when raising it to a power.

    I know that text book publishers think that trig function don't need brackets around their arguments and (even though it is a common abuse of notation) that \cos^2 x is acceptable for (\cos(x))^2 but it really is not.

    CB
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  5. #5
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    Getting back on topic, recall that \displaystyle \tan^2{x} + 1 \equiv \sec^2{x}.

    Your equation is \displaystyle a\tan{x} - b\sec^2{x} - c = 0.

    Substitute \displaystyle \tan^2{x} + 1 where \displaystyle \sec^2{x} is, then you have a quadratic equation you can solve.
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  6. #6
    Newbie SEZHUR's Avatar
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    Quote Originally Posted by Prove It View Post
    Getting back on topic, recall that \displaystyle \tan^2{x} + 1 \equiv \sec^2{x}.

    Your equation is \displaystyle a\tan{x} - b\sec^2{x} - c = 0.

    Substitute \displaystyle \tan^2{x} + 1 where \displaystyle \sec^2{x} is, then you have a quadratic equation you can solve.
    You little beauty! I can finally check my solution, thank you so much.
    Edit: Finally finished incorporating this answer into the question, turns out i was right (about my solution), and wrong (about the course being unlikely to use trig identities). You'll probably be seeing more of me in the future
    Again, thanks.
    Last edited by SEZHUR; March 15th 2011 at 03:27 AM.
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