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 problem

  1. Mar 9, 2006 #1
    Can someone please check over this real quick. I think I’ve solved this problem correctly, but, I’m not sure.

    Solve: [tex]2\sin\theta\cos\theta + 2 \sin\theta - \cos\theta -1 = 0 [/tex]


    [tex]2 \sin\theta(\cos\theta + 1) -(\cos\theta + 1) = 0 [/tex]

    [tex]2 \sin\theta(\cos\theta + 1) = (\cos\theta + 1)[/tex]

    [tex]2 \sin\theta = 1[/tex]

    [tex]\sin\theta = 1/2[/tex]

    [tex]\theta = (\frac{\pi}{6} + k2\pi)\ \mbox {or}\ (\frac{5\pi}{6} + k2\pi) [/tex]
     
    Last edited: Mar 9, 2006
  2. jcsd
  3. Mar 9, 2006 #2
    Looks good to me. The hard way to check this is to plug your answer into your original equation and do the trig to see if it works. The easy way is to pick several values of k and plug it into your original equation using a calculator. :biggrin:

    -Dan
     
  4. Mar 9, 2006 #3
    OK, thanks for your help!
     
  5. Mar 10, 2006 #4

    VietDao29

    User Avatar
    Homework Helper

    Hmm, this step looks wrong.
    What if [tex]cos \theta + 1 = 0[/tex]? And you cannot divide both sides by 0!!!
    The better way is to factor out [tex]cos \theta + 1[/tex], like this:
    [tex]2 \sin\theta(\cos\theta + 1) -(\cos\theta + 1) = 0 [/tex]
    [tex]\Leftrightarrow (\cos \theta + 1)(2 \sin\theta - 1) = 0 [/tex]
    Now if the product of them is 0, then either [tex]\cos \theta + 1 = 0[/tex] or [tex]2 \sin\theta - 1 = 0[/tex], right?
    Can you go from here? :)
    --------------
    @ topsquark, it's not correct, bud. :smile:
     
  6. Mar 10, 2006 #5
    Oh, I see your point. That slipped by me. Sorry. Good catch, VeitDao29!

    -Dan
     
    Last edited: Mar 10, 2006
  7. Mar 10, 2006 #6
    Thanks for the help VietDao. Can you please show me how you got [tex]\ (\cos \theta + 1)(2 \sin\theta - 1) = 0\ [/tex]? I can’t seem to figure out exactly how you did that.
     
  8. Mar 10, 2006 #7
    Oh wait, nevermind, I see how you got that. Duh.

    So, from there I would get:

    [tex]\cos \theta + 1 = 0[/tex]

    [tex]\cos \theta = -1[/tex]

    [tex]\theta = \pi[/tex]

    OR

    [tex]2 \sin\theta - 1 = 0[/tex]

    [tex]\sin\theta= \frac{1}{2}[/tex]

    [tex]\theta = \frac{\pi}{6} [/tex], [tex]\theta = \frac{5\pi}{6} [/tex]
     
    Last edited: Mar 10, 2006
  9. Mar 10, 2006 #8

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Looks right to me, whats your limits?
     
  10. Mar 10, 2006 #9
    By limits, do you mean, for example [tex] (0, 2\pi)[/tex]?

    If that's the case, the question didn't impose any limits, so I suppose the correct answer would be [tex]\theta= (\pi+k2\pi,\ \frac{\pi}{6}+k2\pi\ or, \frac{5\pi}{6}+k2\pi)[/tex].
     
  11. Mar 10, 2006 #10

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yeah thats what I meant.

    When ever I have solved trig equations I am required to find a quantative answer for [itex]\theta[/itex], but if your tutor is happy with you leaving it in terms of [itex]k[/itex] thats all that matter. :smile:
     
  12. Mar 10, 2006 #11
    Thanks for your Help!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Trig problem
  1. Trig problem (Replies: 1)

  2. Trig Problem (Replies: 1)

  3. Trig problem (Replies: 14)

  4. Trig problems (Replies: 7)

  5. Trig problem (Replies: 6)

Loading...