How to plot the graph of #f(x)=cos (pi/2)#?

1 Answer
Jun 26, 2015

Answer:

I suspect you forgot...#x#!
It is probably #f(x)=cos(pi/2x)#
As it is the function represents a constant (#cos(pi/2)=0#), i.e. a horizontal line.

Explanation:

If it is #f(x)=cos(pi/2x)#, you have a cosine in the form:
#f(x)=Acos(kx)#;
of amplitude #A=1# and period equal to: #period=(2pi)/k=(2pi)/(pi/2)=4#:
Graphically you have for #f(x)=cos(pi/2x)#:
graph{cos((pi/2)x) [-5.55, 5.547, -2.773, 2.774]}

As you can see one complete oscillation fits between #0# and #4# radians and then it repeats itself again (period). The maximum height is #1# (amplitude).

So, basically, it is a "shrunk" version of a normal #cos# (that completes one oscillation in #2pi=6.28# radians) that you can see in the following graph of #f(x)=cos(x)#:
graph{cos(x) [-5.55, 5.547, -2.773, 2.774]}