How do you graph \(\sin \left(2 x-\frac{\pi}{2}\right) ?\)
Answer 1:
= 2cos (2x)
Explanation:
Use trig unit circle and property of complement arcs:
y = – 2sin (2x – pi/2) = 2cos (2x)
Answer 2:
See below for step-by-step analysis and graphing
Explanation:
Let’s start with the standard
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 \([0,2 \pi]\)
<><><><><><><><><><><><><><><><><><><><><><>
Now let’s consider what happens with \(\sin \left(x-\frac{\pi}{2}\right)\)
The basic cycle still requires an argument range of \([0,2 \pi]\), that is \(\left(x-\frac{\pi}{2}\right) \in[0,2 \pi]\)
but for this to be possible, \(x \in\left[\frac{\pi}{2}, \frac{5 \pi}{2}\right]\)
That is the basic cycle will appear to have shifted to the right by \(\frac{\pi}{2}\)
<><><><><><><><><><><><><><><><><><><><><><>
Next let’s consider what happens when we double the value of the variable x causing the argument expression to become \(\left(2 x-\frac{\pi}{2}\right)\)
The basic cycle still needs an argument range \([0,2 \pi]\) that is \(\left(2 x-\frac{\pi}{2}\right) \in[0,2 \pi]\)
which, as we have seen, implies \(2 x \in\left[\frac{\pi}{2}, \frac{5 \pi}{2}\right]\)
which, to be possible, means \(x \in\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right]\)
The horizontal distance from the origin on the X-axis seems to have been “squeezed” by a factor of \(\frac{1}{2}\)
<><><><><><><><><><><><><><><><><><><><><><>
Finally, what happens to our basic cycle when we multiply the entire expression by \((-2) \text { to get } y=-2 \sin \left(2 x-\frac{\pi}{2}\right) ?\)
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).
<><><><><><><><><><><><><><><><><><><><><><>
Of course, this only gives us the basic cycle. For the full graph this cycle is repeated infinitely in both directions:
