How do you find the period of #y = 4sin (1/2)(x + pi/4)#?

1 Answer
J'Neal
Nov 2, 2016

By getting your equation in the form #Asin(Bx+C)# and solving #(2pi)/|B|#

Explanation:

First of all, you want to get your equation in the form of #(Bx+C)#. it's not necessary in this case, but just for the sake of learning and avoiding mistakes, I'm including it anyway. You'll want to multiply #(1/2)# and #(x+pi/4)# to get the equation in the form #Asin(Bx+C)# as #4sin(1/2 x+ pi/8)#. Now, we are certain that B equals #1/2#. Next, we'll plug in to to #(2pi)/|B|# #-># #(2pi)/(1/2)# = #4pi#
The period equals #4pi#

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