## Finding the phase shift

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
cruxxfay
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### Finding the phase shift

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

nona.m.nona
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Joined: Sun Dec 14, 2008 11:07 pm
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### Re: Finding the phase shift

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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