1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

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

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    Try plugging in other values too. ...notice anything?:wink:
     
  13. Dec 18, 2008 #12
    sorry, yep your right!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Trig identity question
  1. Trig Identity Question (Replies: 3)

  2. Trig identity questions (Replies: 28)

Loading...