Finding the phase shift  TOPIC_SOLVED

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.

Finding the phase shift

Postby cruxxfay on Wed Jan 02, 2013 5:25 am

I understand that in asin(bx + c) + d, where c represents the phase shift. But then in a question like:

f(x) = 7sin[6(x + pi/12)] + 3

the phase shift here is pi/12 (I thought it was pi/2)

and

f(x) = -3sin(2x + pi/2) - 5

the phase shift here is pi/4...again I thought it was pi/2. So how exactly do you find the phase shift...?

NOTE: I know how to sketch the graph of such, the question, though, is not asking me to sketch the graph, it is simply asking for the phase shift.

Thanks in advance
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Re: Finding the phase shift  TOPIC_SOLVED

Postby nona.m.nona on Wed Jan 02, 2013 4:16 pm

cruxxfay wrote:I understand that in asin(bx + c) + d, where c represents the phase shift. But then in a question like f(x) = 7sin[6(x + pi/12)] + 3, the phase shift here is pi/12 (I thought it was pi/2), and [in] f(x) = -3sin(2x + pi/2) - 5, the phase shift here is pi/4...again I thought it was pi/2. So how exactly do you find the phase shift...?

The process is mentioned here. In general, this may be thought of as finding the value necessary to re-set the argument to the standard starting point, being zero. For f(x) = 7sin[6(x + pi/12)] + 3, the value of x necessary to set 6(x + pi/12) equal to zero is x = -pi/12. Thus, the phase shift is pi/12 (to the left). For f(x) = -3sin(2x + pi/2) - 5, the value of x necessary to set 2x + pi/2 equal to zero is 0 = 2x + pi/2, -pi/2 = 2x, or -pi/4 = x (again, to the left).
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