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Behaviour Under Simple Transformations

  • Thread starter haleym
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  • #1
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Homework Statement


Find the equation of a sine function that has a vertical displacement 2 units down, a horizontal phase shift 60 degrees to the right, a period of 30 degrees, a reflection in the y-axis and an amplitude of 3.

2. The attempt at a solution
Y = 3 sin (30x- 60) -2

I'm not completely sure how to demonstrate a reflection or how to calculate the period...
 

Answers and Replies

  • #2
verty
Homework Helper
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Reflection in the y axis just means that what is positive becomes negative. If f(x) was 2, it becomes -2. So just multiply through by -1.

The period is given by the number in front of x. If sin(x) has a period of 2pi or 360 degrees, sin(kx) has a period of 2pi/k or 360/k degrees. So for a period of 30 degrees, 360/k = 30, for example, or in radians, 2pi/k = pi/6.

Actually, now that I think about it, one would probably always say that a sine function has a period of 360 degrees because it represents a rotation. A rotation is always a whole rotation. I don't have a clear conception of how one would describe the period of a sinusoidal function, exept to refer to whatever coordinates are being used. If x represents degrees (and sin expects its argument in degrees), then sin(2x) has a period of 180 degrees, even though it sounds strange to say that.
 
Last edited:
  • #3
eumyang
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Be careful about the phase shifts, too. Because the phase shift of
Y = 3 sin (30x- 60) -2,
as you wrote it (ignoring for a moment that the 30 in front of the x is wrong) is not 60, but 2.

The generic form for a sinusoid (as I learned it) is
[tex]f(x) = a\sin (b(x - h)) + k[/tex]
where:
|a| indicates a vertical stretch/shrink,
2π/b represents the period,
h represents the phase shift, and
k represent the vertical translation.

So what you have is
Y = 3 sin (30x- 60) -2 = 3 sin(30(x - 2)) - 2,
so the phase shift here is actually 2.
 
  • #4
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Thank you!
 

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