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Math Help - [SOLVED] Trig

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    Question [SOLVED] Trig

    can some one please show me how to show that this equation is true?

    cot^2x-cos^2x= cot^2x * cos^2x

    and i have one more question:
    in solving:
    4cos^2x-3=0
    i get +- sqrt. 3/2
    this means that there are 4 answers, but in the book, it gives me 2 :
    pi/6+npi and 5pi/6+npi
    why?


    thanks
    Last edited by >_<SHY_GUY>_<; May 6th 2008 at 04:07 PM. Reason: missing info
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    can some one please show me how to show that this equation is true?

    cot^2x-cos^2x= cot^2x * cos^2x
    Consider the LHS.

    \cot^2x - \cos^2 x = \frac {\cos^2 x}{\sin^2 x} - \cos^2 x

    = \cos^2 x \left( \frac 1{\sin^2 x} - 1 \right)

    = \cos^2 x (\csc^2 x - 1)

    = \cos^2 x \cot^2 x


    and i have one more question:
    in solving:
    4cos^2x-3=0
    i get +- sqrt. 3/2
    this means that there are 4 answers, but in the book, it gives me 2 :
    pi/6+npi and 5pi/6+npi
    why?


    thanks
    no, there are infinitely many answers! you were not given any bounds. the answers given are the general formulas for ALL solutions
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  3. #3
    Member >_<SHY_GUY>_<'s Avatar
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    so that would include the solutions in quads 3 and 4?
    what is LHS?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    so that would include the solutions in quads 3 and 4?
    because of the +/-, you will have solutions in all quadrants

    \frac {\pi}6 + n \pi takes care of all solutions in quadrants 1 and 3, while \frac {5 \pi}6 + n \pi takes care of all solutions in quadrants 2 and 4
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  5. #5
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    so that would include the solutions in quads 3 and 4?
    what is LHS?
    LHS = Left-hand side

    And yes, there are solutions in the 3rd and 4th quadrants. Notice the forms of the answers are given as x + \pi n and not x + 2\pi n.
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  6. #6
    Member >_<SHY_GUY>_<'s Avatar
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    so that would explain that it would alternate to quad 1 to quad 3...solution wise?

    how do you get that equation to look like that?
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