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Trig identities

  1. Jul 7, 2009 #1

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

    2sin(2x) - [tex]\sqrt{3}[/tex] = 0

    Find the value of the variable.


    2. Relevant equations

    Sin(2x) = 2SinxCosx



    3. The attempt at a solution
    I subtracted root3 and then divided everything by 2 which leaves me with

    sin (2x) = [tex]\sqrt{3}[/tex]/2

    then I used the double angle identity to change sin 2x into 2sinxcosx.

    I'm stuck from there. I forgot how to use double angles ( eek.)
     
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  3. Jul 7, 2009 #2

    jgens

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    I wouldn't use the double angle identity to solve this problem. Go back to, sin(2x) = (3)1/2/2 and take the inverse sine.
     
  4. Jul 7, 2009 #3

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    How do I do that?
     
  5. Jul 7, 2009 #4

    jgens

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    If asked you to find all x that satisfy, sin(x) = 1/2, what would you do? If you can do that problem, then you can do this one.
     
  6. Jul 7, 2009 #5

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    I'm still stuck. I used the inverse sin of the (root3)/2 and I got answers like pi/3, 2pi/3, ect.

    I think I need to use the identity but it seems impossible to figure out:

    2sinx cosx = (root3)/2
     
  7. Jul 7, 2009 #6

    jgens

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    You don't need the identity. You got multiple answers because the trigonmetic functions are periodic so there are an infinite number of solutions to equations like sin(x) = 1/2. You really want to use the inverse sin in this case.
     
  8. Jul 7, 2009 #7

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    What happens to the 2x ?
     
  9. Jul 7, 2009 #8

    jgens

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    You end up solving it for x, just the same as if I told you 2x = 6.
     
  10. Jul 7, 2009 #9

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    HOLY oh man, I am real angry right now.

    I left my calculator in radian mode and I was using degrees for like the last 40 mins !
     
  11. Jul 7, 2009 #10

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    Whats significance does the identity: sin2x = 2sinxcosx have?
     
  12. Jul 7, 2009 #11

    jgens

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    It doesn't have significance in this problem. It's useful when you need to manipulate trigonometric functions to put them into an integrable form or when you need to prove other identities.
     
  13. Jul 7, 2009 #12
    let t=2x.

    can you solve sin(t) = sqrt(3)/2
     
  14. Jul 7, 2009 #13

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    yeah, its pi/6, 5pi/6, and so forth.
    Whats the sign that I add to the answer if it's periodical? How exactly do I write the answer for this problem? Do I write all the radians that make the equation true one time around the unit circle then add the (periodical)?
     
  15. Jul 8, 2009 #14
    You usually write your answers with k where k is any integer; for example, solutions for sin(x) = 1 is x = pi/2 + 2k*pi
    Your answer will need a little more than just 2k*pi
     
    Last edited: Jul 8, 2009
  16. Jul 8, 2009 #15
    You should end up with multiple answers as said before trig functions are periodical.
    Did the question have any limits? i.e 0<x<2pi/360??
     
  17. Jul 8, 2009 #16

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    No limits.

    So my answer is basicaly, 30 degrees or pi/6 + 2kpi? Correct?
     
  18. Jul 8, 2009 #17
    pi/6 + k*pi takes care of half of the answers; you also need the solutions with 5pi/6.
     
    Last edited: Jul 8, 2009
  19. Jul 8, 2009 #18
    [tex]sin(2x)=\frac{\sqrt{3}}{2}[/tex]

    [tex]2x=\begin{bmatrix}
    \frac{\pi}{3} + 2k\pi\\
    \pi-\frac{\pi}{3}+2k\pi
    \end{bmatrix}[/tex]

    Can you solve it now?
     
  20. Jul 8, 2009 #19

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    x = pi/6 + 2kpi, 5pi/6 = 2kpi?

    On a test or something, would I just leave the answers as so? ( if their correct).
     
  21. Jul 8, 2009 #20

    Mark44

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    No.
    Starting with Дьявол's work (which I haven't checked, but believe is correct), you have
    2x = pi/3 + 2k*pi, or 2x = 2pi/3 + 2k*pi, so
    x = pi/6 + k*pi, or x = pi/3 + k*pi, where k is any integer.

    You forgot to divide the 2k*pi parts.
     
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