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Homework Help: Trig identity question

  1. Dec 17, 2008 #1
    1. The problem statement, all variables and given/known data

    Basically I am finishing of a projectile question and I get stuck here:

    Trying to find [itex] \theta [/itex]

    [tex] \frac{1}{2} (sin2 \theta) tan^2 \theta -tan \theta + \frac{1}{2} sin2 \theta = 0[/tex]

    2. Relevant equations



    3. The attempt at a solution
    I tryed spliting [itex] tan \theta [/itex] into [itex] \frac{sin \theta}{cos \theta} [/itex] but I don't really get anywhere.
    I know it requires using a trig identity but I can't really see a suitable one.
    Thanks for the help in advance!!
     
  2. jcsd
  3. Dec 17, 2008 #2

    gabbagabbahey

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    The identities [itex]\sin(2\theta)=2\sin(\theta)\cos(\theta)[/itex] and [itex]\sec^2(\theta)=\tan^2(\theta)+1[/itex] should both be useful.

    However, you won't be able to solve your equation for [itex]\theta[/itex] since it turns out your equation is true for all [itex]\theta[/itex]!:eek:

    You must have made an error earlier in the problem.
     
  4. Dec 17, 2008 #3
    hmm I got sin x =0 so x=0.
     
  5. Dec 17, 2008 #4
    [tex] sin \theta cos \theta (tan^2 \theta) - tan \theta + sin \theta cos \theta = 0 [/tex]

    [tex] sin \theta cos \theta (sec^2 \theta) - \frac{sin \theta}{cos \theta} = 0 [/tex]

    [tex] cos^2 \theta (sec^2 \theta) = 1 [/tex]

    [tex] 1 =1 [/tex]

    ?
     
  6. Dec 17, 2008 #5

    gabbagabbahey

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    Then you should double check your algebra :wink:
     
  7. Dec 17, 2008 #6

    gabbagabbahey

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    Like I said; your equation is true for all [itex]\theta[/itex], so you must have made an error earlier in the problem.
     
  8. Dec 18, 2008 #7
    First thing I did was find rewrite everything in terms of sin and cos:

    [tex] sin(\theta)cos(\theta) \frac{sin^{2}(\theta)}{cos^{2}(\theta)} - \frac{sin(\theta)}{cos(\theta)} + sin(\theta)cos(\theta) = 0 [/tex]

    Then I found the common denominator and got:

    [tex] \frac{sin^{3}(\theta) - sin(\theta) + sin(\theta)cos^{2}(\theta)}{cos(\theta)} = 0 [/tex]

    So now we have to consider 2 cases

    1) [tex] cos(\theta) = 0 [/tex]

    That case cannot hold because your original expression has tan in it and therefore it is assumed that [tex] cos(\theta) \neq 0 [/tex]

    2) [tex] cos(\theta) \neq 0 \Rightarrow sin(\theta)(sin^{2}(\theta) - 1 + cos^{2}(\theta)) = 0 [/tex]

    The 2nd case gives you 0 = 0 so it seems like your expression holds true for any x.
     
  9. Dec 18, 2008 #8
    try searching you double angle rules for something related to sin2x=2sinxcosx and apply it
     
    Last edited: Dec 18, 2008
  10. Dec 18, 2008 #9
    lol but if you put zero degree in you do get 0. I have probability done something wrong.
     
  11. Dec 18, 2008 #10
    Zero degree? Are you saying if you evaluate it at theta = 0 you get 0 = 0? Did you read the post I made?
     
  12. Dec 18, 2008 #11

    gabbagabbahey

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    Try plugging in other values too. ...notice anything?:wink:
     
  13. Dec 18, 2008 #12
    sorry, yep your right!
     
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