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Trig Solve for Solution Within Interval

  1. Sep 28, 2006 #1
    Hi, I have a problem that says to solve this equation for the interval 0<theta<2 pie which is one revolution of the unit circle. The equation is:

    (square root of 2)(cos 2theta) = 1

    The cos2theta is a double angle, but I couldn't find any solutions for theta that would = 1 at the end. The answer has to fit in for the theta angle. Is this possibly no solution? I had originally put pie/4 as an answer but realized I forgot the parenthesis in the calculator and now I need to redo the problem. Thanks!
     
  2. jcsd
  3. Sep 28, 2006 #2
    Draw the unit circle in the x-y cartesian plane. If you rotate around the circle by an angle [itex]\theta[/itex] counterclockwise wrt the positive x-axis, then [itex]\cos{\theta}[/itex] is the x-coordinate of the resulting point. So what [itex]\theta[/itex] will make [itex]\cos{\theta} = \frac{1}{\sqrt{2}}[/itex] (HINT: What's [itex]2(1/\sqrt{2})^2[/itex]?)?

    Edit: courtrigrad I suggest plugging [itex]\pi / 2[/itex] into [itex]\sqrt{2}\cos{2\theta}[/itex] and seeing what you get :tongue2:
     
    Last edited: Sep 28, 2006
  4. Sep 28, 2006 #3
    my fault it should have been [tex] \theta = \frac{\pi}{8} [/tex]. I multiplied instead of divided.
     
  5. Sep 28, 2006 #4
    You can give a closed form for [itex]\theta[/itex].

    Edit: :smile:
     
  6. Sep 28, 2006 #5
    [tex] \theta = \frac{\pi}{8} + 2n\pi [/tex]
     
  7. Sep 28, 2006 #6
    he said [itex] 0 < \theta < 2\pi[/itex] though (of course, there's still [itex]\theta = \frac{15}{8} \pi[/itex]!) :tongue:
     
    Last edited: Sep 28, 2006
  8. Sep 28, 2006 #7
    Thank you both so much for the help! I really appreciate it and understand now.
     
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