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Trig sin function Question

  1. Jan 29, 2008 #1
    Greetings,

    I've solved (almost) a problem - with the answer involving sin().

    My first solution involves values for which sin() is equal to 0.5 - i.e.30 degrees or 0.5235.

    The internal of the sin function is sin(10t -0.927)
    t = 0.1451

    To complete this solution I need to show the other solutions for this - that is the other values of t for which sin (10t - 0.927) = 0.5.

    I thought it would be every pi/2 but I see that does not work.

    It's obvious less than pi and greater than pi/2 - meaning the simple sin wave starts at 0 goes through 0.5 crests at 1 at pi/2 and goes back through 0.5 (at what value?) and to 0 at pi.

    Thanks so much
    -Sparky_
     
  2. jcsd
  3. Jan 29, 2008 #2

    mathman

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    Science Advisor
    Gold Member

    sin(pi/6)=sin(5pi/6)=.5

    The general form is sinx=sin(pi-x).

    I can't figure out the rest of your post.
     
  4. Jan 29, 2008 #3
    I was solving sin(10t-0.927) = 0.5
    t = 0.1451

    but it will actually be t = 0.1451 + (something periodic)

    how do I find the something periodic? - meaning t = 0.1451 =n*pi/2 or some such?
     
  5. Jan 29, 2008 #4
    Suppose you set 10t-0.927 equal to a new variable y. Then you would have sin(y) = 0.5, and

    [tex]y=\frac{\pi}{6} + 2n\pi \text{ or }y=\frac{5\pi}{6} + 2n\pi[/tex]

    because sine has a period of [tex]2\pi[/tex], and [tex]\sin a = \sin(\pi - a)[/tex]. Replace y with 10t-0.927 in both of those cases and solve for t.
     
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