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Simple trigonometric problem

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex] \cos (2 \theta)=- \cos ( \theta)[/tex]


    2. Relevant equations



    3. The attempt at a solution
    In general [tex] \theta = 2n \pi +/- \alpha \mbox{ where } \cos \theta = \cos \alpha [/tex]
    Because [tex] \cos(2 \theta) = - \cos( \theta) \mbox{ then, } 2 \theta = 2n \pi -/+ \theta[/tex]
    taking [tex] 2 \theta = 2n \pi - \theta \mbox{ that implies } 3 \theta = 2n \pi \mbox{ therefore } \theta = \frac{2}{3}n \pi [/tex]. The book says that theta must lie between 0 and pi/2. So I must have gone wrong. Can anyone put me right on this equation. Thanks for the help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 26, 2010 #2
    This is confusing. Are you trying to show that the left-hand side is equal to the right-hand side (which I don't think is true) or are you trying to solve some other problem?
     
  4. Apr 26, 2010 #3
    I am trying to show that the left side equals the right, but I am not sure if I am correct. The equation itself results from solving another problem. Thanks.
     
  5. Apr 26, 2010 #4

    ehild

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    Homework Helper
    Gold Member

    cos(theta) = -cos(alpha) implies that either

    theta = alpha +2n*pi

    or

    theta= (pi-alpha) +2n*pi.

    ehild
     
  6. Apr 26, 2010 #5
    You made a mistake.

    [tex]-cos\theta=cos(\pi+\theta)[/tex]
    [tex]2\theta_{n,+}=\left(2n+1\right)\pi+\theta_{n,+}[/tex]
    [tex]2\theta_{n,-}=\left(2n+1\right)\pi-\theta_{n,-}[/tex]
    [tex]\theta_{n,+}=\left(2n+1\right)\pi[/tex]
    [tex]\theta_{n,-}=\frac{\left(2n+1\right)\pi}3[/tex]
    There are no solutions [tex]\theta_{n,+}[/tex] within the interval [tex]\left[0,\frac \pi 2\right][/tex], but the solution [tex]\theta_{0,-}=\frac \pi 3[/tex] exists.

    (Edit: Yeah, what they said.)
     
  7. Apr 26, 2010 #6
    No, the equation I want solved is [tex] \cos (2 \theta)= - \cos( \theta)[/tex]. I do not know how to do it. Thanks.
     
    Last edited: Apr 26, 2010
  8. Apr 26, 2010 #7
    Thanks Gigasoft, and thank you all.
     
  9. Apr 27, 2010 #8

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What you need is the identity [itex]cos(2\theta)= cos^2(\theta)- sin^2(\theta)=[/itex][itex] cos^2(\theta)- (1- cos^2(\theta))[/itex]

    [itex]cos(2\theta)= 2cos^2(\theta)- 1[/itex].

    You get a quadratic equation in [itex]cos(\theta)[/itex].
     
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