How do you graph #sin(2x-pi/2)#?

Use trig unit circle and property of complement arcs:
y = - 2sin (2x - pi/2) = 2cos (2x)

Let's start with the standard #color(black)(sin(x))# function
and in particular let's note the basic cycle of the #sin(x)# function:

Notice that the argument range for the basic cycle is # [0,2pi]#

#color(magenta)("<><><><><><><><><><><><><><><><><><><><><><><><><><><> ")#

Now let's consider what happens with #sin(x-pi/2)#
The basic cycle still requires an argument range of #[0,2pi]#,
that is #(x-pi/2) in [0,2pi]#
but for this to be possible, #x in [pi/2,(5pi)/2]#

That is the basic cycle will appear to have shifted to the right by #pi/2#

#color(magenta)("<><><><><><><><><><><><><><><><><><><><><><><><><><><> ")#

Next let's consider what happens when we double the value of the variable #x# causing the argument expression to become #(2x-pi/2)#
The basic cycle still needs an argument range #[0,2pi]#,
that is #(2x-pi/2) in [0,2pi]#
which, as we have seen, implies #2x in [pi/2,(5pi)/2]#
which, to be possible, means #x in [pi/4,(5pi)/4]#

The horizontal distance from the origin on the X-axis seems to have been "squeezed" by a factor of #1/2#

#color(magenta)("<><><><><><><><><><><><><><><><><><><><><><><><><><><> ")#

Finally, what happens to our basic cycle when we multiply the entire expression by #(-2)# to get #y=-2sin(2x-pi/2)#?
The value of each y-coordinate is scaled by a factor of #2#
(you might think of this as the coordinate is reflected in the X-axis and then stretched by #2# away form the X-axis).

#color(magenta)("<><><><><><><><><><><><><><><><><><><><><><><><><><><> ")#

Of course, this only gives us the basic cycle.

For the full graph this cycle is repeated infinitely in both directions: