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How would I solve sinѲ=sin2Ѳ?

  1. Sep 1, 2014 #1
    1. The problem statement, all variables and given/known data

    sinΘ=sin2Θ

    2. Relevant equations

    Double-angle forumla...
    sin2Θ=2sinΘcosΘ

    sinΘ=2sinΘcosΘ
    sinΘ

    3. The attempt at a solution
    sinΘ=2sinΘcosΘ
    sinΘ-2sinΘcosΘ=0
    sinΘ(1-2cosΘ)=0

    Now I'm getting stuck because 1-2cosΘ is not an identity of any sort.

    Any help would be appreciated...thanks!
     
  2. jcsd
  3. Sep 1, 2014 #2

    Nathanael

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    Instead of subtracting; what about dividing both sides by sinΘ?


    Edit:
    You could divide right here, too. (zero divided by anything* is still zero)


    *anything except zero
     
  4. Sep 1, 2014 #3
    I'll try that!
    Edit:
    AH I see it
     
  5. Sep 1, 2014 #4

    Mark44

    Staff: Mentor

    This is not a good idea. Dividing by sinθ causes you to lose solutions for which sinθ = 0.
     
  6. Sep 1, 2014 #5

    Nathanael

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    Actually, your way is better
    It is better to set it up like this because this shows all of the solutions. (I overlooked this in my first post.)


    When is sinΘ(1-2cosΘ) equal to zero? (There are two solutions)



    edit:
    Yes I just realized this :redface: thank you




    edit#2:
    I had a feeling you would point this out too :biggrin: I couldn't think of a concise way to say it. (I meant two ways)
    Maybe I should stay out of the math section :tongue:
     
    Last edited: Sep 1, 2014
  7. Sep 1, 2014 #6

    Mark44

    Staff: Mentor

    Doesn't need to be. You're on the right track. If the product above equals zero, that means that either sinθ = 0 or 1 - 2cosθ = 0.

     
  8. Sep 1, 2014 #7

    Mark44

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    There are LOTS more than two solutions...
     
  9. Sep 1, 2014 #8

    nrqed

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    You have the product of two functions of Θ. The result is zero when either sinΘ or 1-2cosΘ is zero.
    This occurs at what angles? (As for 1-2cosΘ=0, just isolate the angle. Do you see how to do that?)
     
  10. Sep 1, 2014 #9
    Let's see...
    I could do sinΘ=0 or 1-2cosΘ=0
    sinΘ=0
    Θ= (pi)/2
    1-2cosΘ=0
    2cosΘ=-1
    cosΘ=-1/-2
    cosΘ=1/2
    Θ=5(pi)/3 or (pi)/3
    I see that if I do it this way, instead of dividing out the sine, I capture one more solution.
    Very helpful...thank you!
     
  11. Sep 1, 2014 #10

    Mark44

    Staff: Mentor

    Since there are no restrictions on θ (or at least none shown in your OP), you should add integer multiples of 2π to each of your solutions to capture all possible solutions.
     
  12. Sep 1, 2014 #11

    nrqed

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    Watch out, ## \sin(\theta) =0 ## at ##\theta = \ldots, - \pi, 0 , \pi, \ldots ##,

    not at ##\theta = \pi/2##!
     
  13. Sep 2, 2014 #12

    BvU

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    Watch out some more !
    sinΘ(1-2cosΘ)=0 ##\Leftrightarrow ## sinΘ=0 ##\vee ## (1-2cosΘ)=0

    First one: sinΘ=0 ##\Leftrightarrow ## sinΘ=sin(0) ##\Leftrightarrow## Θ = 0 + 2n∏ ##\vee ## Θ = ∏ - 0 + 2n∏ ##\Leftrightarrow## Θ = n∏ (with n an integer)

    Now you do the second one. And don't write 1-2cosΘ=0 on one line and 2cosΘ=-1 on the next !

    Being meticulous while learning this pays off all through the rest of your life (and that of possible pupils/students you might have later on :smile:)

    By the way, welcome to PF !
     
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