How do you find the amplitude and period for #y=3sin2 (x-pi/6)#?

1 Answer
Jun 19, 2015

Answer:

Period: #pi#
Amplitude: #6#

Explanation:

We start analyzing the argument of sin in #y=3sin2 (x-pi/6)#:
If the function was #y=sinx# we would have a period of #2pi# because the sine function makes a complete oscillation (0,1,0,-1,0) in #2pi#.
If we sum anything to the argument (for example like the function #y=sin(x+999pi+sqrt(7))#), we make a horizontal translation, so we won't change neither the period or the amplitude.
So we can consider a similar function with the same amplitude and period that is easier to study: #y=3sin2x#.

Amplitude
If we consider that the sine function has a range #[-1,1]#, its amplitude is 2. If the argument varies in all #RR#, the range won't change. For example #y=sin(sinx+34x)# would have always the same amplitude (2) and range ([-1,1]).
What really matters in amplitude is the vertical dilation/compression factor outside of sin(x): in this case we would have #y=3sinx# that means a range of #[-3,3]# so an amplitude of #3-(-3)=6#.

Period
If we consider the sine function, it has a period of #2pi#. The cohefficient inside the argument is a horizontal dilation/compression of the function. We could find the value of the period deviding #2pi# by this cohefficient. For example in this case we should have #(2pi)/2=pi#.

Verify
As we can see in this graph, the range is [-3,3], and the function is plotted in the interval #[0,pi]#. Zooming out, we can see that it's periodic and it's obviously translated.

graph{3sin(2(x-pi/6)) [0, 3.14, -3, 3]}