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.