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Math Help - Trigonometric Identities, need help fast!

  1. #1
    Newbie
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    Nov 2008
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    Trigonometric Identities, need help fast!

    I'm reading on a distance, and this is the first time I just got totally stuck. I need help really fast!

    It's probably really easy.

    I understand that sin^2v + cos^2v = 1

    Now for an example:
    _________________________________________
    What is Sin(v) if cos(v) = -(24/25)?
    I'm doing sin^2(v) = 1-(-(24/25)^2) = 1.9216 which ofcourse is wrong (answer is -(7/25)).
    _________________________________________

    Example 2:

    _________________________________________
    1/cos^2(v) = 1 + tan^2(v)
    This is how the solution is written in the book:
    1+tan^2(v) = 1 + (sin^2(v)/cos^2(v)) = (cos^2(v) + sin^2(v))/cos^2(v)

    For some reason I don't get how 1 becomes cos^2(v).
    _________________________________________

    What am I doing wrong? Appreciate help!!
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  2. #2
    MHF Contributor
    Joined
    Mar 2007
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    1,240

    Talking

    Quote Originally Posted by vane505 View Post
    What is Sin(v) if cos(v) = -(24/25)?
    I'm doing sin^2(v) = 1-(-(24/25)^2) = 1.9216 which ofcourse is wrong
    Be a little more careful with your "minus" signs. If cos(v) = -24/25, then [cos(v)]^2 = [-24/25]^2, not -[24/25]^2. The square is on all of the value, not just the fraction!

    Quote Originally Posted by vane505 View Post
    1/cos^2(v) = 1 + tan^2(v)
    This is how the solution is written in the book:
    1+tan^2(v) = 1 + (sin^2(v)/cos^2(v)) = (cos^2(v) + sin^2(v))/cos^2(v)
    For some reason I don't get how 1 becomes cos^2(v).
    It wasn't that the 1 "became" a cosine squared. It's that they converted the "1" to the common denominator:


    . . . . . 1\, +\, \frac{\sin^2{(v)}}{\cos^2{(v)}}

    . . . . . \frac{\cos^2{(v)}}{\cos^2{(v)}}\, +\, \frac{\sin^2{(v)}}{\cos^2{(v)}}

    . . . . . \frac{\cos^2{(v)}\, +\, \sin^2{(v)}}{\cos^2{(v)}}

    Hope that helps!
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  3. #3
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    Thank you! Got both of them now
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