Problem 20 - Algebra and Trigonometry

Question

Determine the amplitude, period, and phase shift of each function. Then graph one period of the function. $$y=\sin \left(2 x-\frac{\pi}{2}\right)$$

Step-by-Step Solution

Answer

The amplitude of the function \(y = \sin (2x - \frac{\pi}{2})\) is 1, the period is \(\pi\) and the phase shift is \(\frac{\pi}{2}\) to the right.

Step 1: Identifying coefficients in the function

From the function \(y = \sin(2x - \frac{\pi}{2})\), one can see that the coefficient \(a = 1\), \(b = 2\) and \(c = \frac{\pi}{2}\). Here, \(a\) is the amplitude, \(b\) determines the period and \(c\) determines the phase shift of the function.

Step 2: Computing the amplitude

The amplitude of the function is given by the absolute value of coefficient \(a\). So for the given function, the amplitude is \(|1|\), which equals 1.

Step 3: Computing the period

The period \(P\) of the function is given by \(P = \frac{2\pi}{|b|}\). Substituting \(b = 2\), we have \(P = \frac{2\pi}{2}\), thus the period is \(\pi\).

Step 4: Computing the phase shift

The phase shift is given by \(c\). For the given function, \(c = \frac{\pi}{2}\) so the phase shift is \(\frac{\pi}{2}\). We observe that the phase shift is subtracted in the function, which means the graph shifts to the right by \(\frac{\pi}{2}\).

Step 5: Graphing the function

Not achievable here but normally one would draw a sine wave with amplitude of 1, period of \(\pi\) and shifted to the right by \(\frac{\pi}{2}\).