How do you determine the period for #y = cos (2pix)#?

2 Answers
Apr 27, 2015

You can use the #2pi# in front of #x# in the argument of #cos#.
Calling this number #k# you have that:
#k=(2pi)/(period)#
So in your case you have #period=1#
Your function does an entire oscillation in 1 radian.
graph{cos(2pix) [-2.433, 2.435, -1.215, 1.217]}

Apr 27, 2015

Another way to think about it:

#cos(theta)# has a period of #2pi#
You can think of this as meaning that the values
#" from " cos(0) " to " cos(2pi)#
make up one period.

So the question is what values of #x#
would cause #cos(2pi*x)# to take on the range
#cos(0) " to " cos(2pi)#

The answer is obviously #x = 0 " to " 1#
So the period is #1#