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Solving trig function (possible error in solutions)

  1. Nov 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Alright, so I am going through the solutions to one problem that has me stumped, and initial instinct tells me it is wrong, however, I figured I would get clarification.


    3. The attempt at a solution
    [tex]sin(2t)=cos(2t)[/tex]
    [tex]tan(2t)=1[/tex]
    [tex]2t=\pi{/}4, 5\pi{/}4[/tex]

    However, the book states both positive and negative values, rather than just positive. It's either not correct, or not clicking. tan(-π/4)=-1, not 1..
     
  2. jcsd
  3. Nov 18, 2011 #2

    I like Serena

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    What is tan(-3pi/4)?
     
  4. Nov 18, 2011 #3

    dynamicsolo

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    Since your text is suggesting that there are positive and negative solutions, you are not restricted to the "principal circle" , [itex] 0 ≤ t < 2{\pi} [/itex] . So you can "go around the circle" as many times as you like:

    [tex] 2t = \frac{\pi}{4} + k \cdot 2{\pi} , \frac{5\pi}{4} + k \cdot 2{\pi} , [/tex]

    with k being any integer.
     
  5. Nov 19, 2011 #4

    SammyS

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    Are you sure the problem isn't, [itex]\sin^2(t)=\cos^2(t)\,?[/itex]
     
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