Trigonometry question

Question

What is the phase shift of this function?

D(x) = 3sin(2π/365)(x-78)+12

Answer 1

D(x) = 3 sin[(2π/365)x - 156π/365) + 12

Phase shift is -c / b

(156π/365)(36/2π)

156π / 2π

78

Answer 2

 the general sine function:

F(t) = Asin(Bt – C) + D

...where |A| is the amplitude, B gives you the period, D gives you the vertical shift (up or down), and C/B is used to find the phase shift.

As written,

D(x) = 3sin(2π/365)(x-78)+12

is a LINEAR function with slope 3 sin(2*pi/356)

Perhaps, you meant:

D(x) = 3sin(2π/365)(x-78)+12

= 3[sin( (2π/365)x -(2π/365)(78))]+12 =

in which case the phase shift is 78

If the parenthesis are not in the right place, then perhaps you meant:

D(x) = 3sin((2π/365)x-78)+12

in which case the phase shift is:

78 divided by 2π/365

= 78*365/2π = 14235/π