# Sinusoidal Characteristics

• Jun 9th 2011, 05:11 AM
starshine84
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!
• Jun 9th 2011, 05:14 AM
Ackbeet
I would multiply out the argument of the sin function. If you're trying to find the period, you can ignore the phase angle.
• Jun 9th 2011, 05:16 AM
starshine84
I'm not sure what you mean by, "multiply out the argument of the sin function."
• Jun 9th 2011, 05:17 AM
Ackbeet
I mean,

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

I don't mean using the sum of angles formula.
• Jun 9th 2011, 05:34 AM
starshine84
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...
• Jun 9th 2011, 05:36 AM
starshine84
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...
• Jun 9th 2011, 05:42 AM
Ackbeet
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.
• Jun 9th 2011, 05:49 AM
starshine84
Awesome, thanks so much for all of your help!!! :)
• Jun 9th 2011, 05:50 AM
Ackbeet
You're very welcome!