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Math Help - Sinusoidal Characteristics

  1. #1
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    Sinusoidal Characteristics

    I am having difficulty with this function:

    y = -2sin (2 ((t+1)/3)

    I know to find the period I need to divide 360 degrees by the absolute value of b, but I can't figure out how to factor the equation properly to find the correct value of b.

    I thought I could multiply both the 2, and the (t+1)/3 by 3 to eliminate the fraction, but that doesn't seem to be right.

    I think I'm probably missing something fairly obvious... the other questions have been so easy!
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  2. #2
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    I would multiply out the argument of the sin function. If you're trying to find the period, you can ignore the phase angle.
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  3. #3
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    I'm not sure what you mean by, "multiply out the argument of the sin function."
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  4. #4
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    I mean,

    2 ((t+1)/3) = 2t/3 + 2/3.

    I don't mean using the sum of angles formula.
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  5. #5
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    Okay, I see what you're talking about. That works perfectly to find the period, great!

    I am able to find the period, amplitude, max value, min value, range, domain, and vertical displacement from that, but am not sure how to find the horizontal shift then...
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  6. #6
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    Would it be one unit to the right?
    (t+1)/3 = (t-(-1))/3

    Does the 3 impact it at all? I'm a little confused on that...
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  7. #7
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    You're on the right track. Think of horizontal shifting this way: suppose I have a function, x^2. It's zero when x = 0, right? Now suppose I shift it horizontally. That always looks like replacing the x with an x-a or x+a. So which is which? Well, if I replace x with x-2 in our function, then I get (x-2)^2. When is that zero? When x = 2. Evidently, then, the x-2 shifting is to the right. If I were to do x+2, then I'd have (x+2)^2, which is zero when x = -2. So that one's shifted to the left.

    This is how I remember which one is which. Hope that works for you.

    In answer to your question, the 3 does not impact the shifting.
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  8. #8
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    Awesome, thanks so much for all of your help!!!
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  9. #9
    A Plied Mathematician
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    You're very welcome!
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