How do you find the amplitude and period of #y=4cos(72pit)#?

Question

1486 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-30T05:09:35

#Amplitude=4#
#Period=1/36#

For a cos (or sin) function of the form

#y=acos(bt)+c#

the period is given by

#Period=(2pi)/b#

and the amplitude is given by a

#Amplitude=a#

In this case

#a=4#
#b=72pi#

Therefore

#Period=(2pi)/(72pi)=1/36#

#Amplitude=4#