How do you find the amplitude and period of #y=3cos(2/3theta)#?

1 Answer
Nov 1, 2016

For #y=acos(btheta)=3cos(3/2 theta)#
#Amplitude = a = 3#
#Period = (2π)/|b| = (2pi)/(2/3) = (6pi)/2 = 3pi#

Explanation:

For #amplitude#, we need the factor #A#,
so for #y=3cos(3/2 theta)#, that would be 3.

For #period#, we must plug-in #b# to #(2π)/|b|# and simplify.
#b=3/2#, so #(2π)/|b| = (2pi)/(2/3)#. To simplify this, we must first multiply the lowest denominator by the numerator #((2pi)*3)/((2/3)*3)# getting #(6pi)/2#, which we can then simplify dividing by 2 #((6pi)/2)/(2/2)# getting #3pi#.