How do you write the equation of a cos function: amplitude=2/3 period=pi/3 phase shift= -pi/3 vert. shift= 5?

2 Answers
Jan 4, 2017

Answer:

Start with the general form:
#y = (A)cos(Bx + C) + D#
Please see the explanation for how you change it.

Explanation:

A is the amplitude:

#y = 2/3cos(Bx + C) + D#

The period is #T = (2pi)/B#

Substitute #pi/3# for T and then solve for B:

#pi/3 = 2pi/B#

#B = 6#

#y = 2/3cos(6x + C) + D#

The phase shift is #phi = -C/B#

Substitute, #-pi/3" for "phi#, 6 for B, and then solve for C:

#-pi/3 = -C/6#

#C = 2pi#

#y = 2/3cos(6x + 2pi) + D#

The vertical shift is #D = 5#:

#y = 2/3cos(6x + 2pi) + 5# Answer

Here is a graph of the above equation:

graph{y = 2/3cos(6x + 2pi) + 5 [-11.25, 11.25, -5.625, 5.625]}

Jan 4, 2017

Answer:

#y = 5 + 2/3 sin (6 t + 2 pi)#

Explanation:

let's write out something really general first

#y - y_o = A sin (omega t + psi)#

Easy ones first:

Amplitude: #A = 2/3#

Vertical Shift upward: #y_o = 5#

Ergo: #y - 5 = 2/3 sin (omega t + psi)# !!

For the period (am assuming you mean temporal, not spatial, period, though the approach is the same), note that #omega = (2 pi )/T = (2 pi )/(pi/3) = 6 #

#implies y - 5 = 2/3 sin (6 t + psi)#

For phase shift #- pi/3#, you need to be careful. If we re-write this as #y - 5 = 2/3 sin (6 (t + bar psi))#, hopefully you will see that #bar psi# represents the actual shift along the t axis.

We want that shift to be #pi/3# to the left so we can say that

#y - 5 = 2/3 sin (6 (t + pi/3))#

#= 2/3 sin (6 t + 2 pi)#

ie we can say that #psi = 2 pi#

That may seem odd but the period of the function is #T = pi/3# so it is a full phase shift