Question

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

Answer 1

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]}