Problem 23 - Algebra and Trigonometry

Question

Determine the amplitude, period, and phase shift of each function. Then graph one period of the function. $$y=\frac{1}{2} \sin \left(x+\frac{\pi}{2}\right)$$

Step-by-Step Solution

Answer

The amplitude is \(\frac{1}{2}\), the period is 1, and the function is phase shifted to the right \(\frac{\pi}{2}\) units.

Step 1: Calculation of Amplitude

Identify the variable A in our provided function \(y=\frac{1}{2} \sin \left(x+\frac{\pi}{2}\right)\). This is denoted by the coefficient of sin, which in our case is \(\frac{1}{2}\). The amplitude is then the absolute value of A, or \(|\frac{1}{2}|\), which equals \(\frac{1}{2}\).

Step 2: Calculation of the Period

The period is found by taking the reciprocal of the absolute value of B where B is the coefficient of x. Since there is no explicit coefficient in x in our function, we understand that B is 1, and thus the period is \(|1|^{-1} = 1\).

Step 3: Calculation of Phase Shift

The phase shift can be calculated as \(-C/B\). Here, C is the term that is added or subtracted inside the sine function. In our case, C = \(\frac{\pi}{2}\). So, our phase shift is given by \(-C/B = -\frac{\pi}{2} / 1 = -\frac{\pi}{2}\). This means the graph will shift to the right by \(\frac{\pi}{2}\) units.

Step 4: Graphing the Function

To graph the function \(y=\frac{1}{2} \sin \left(x+\frac{\pi}{2}\right)\), we start at the phase shift, which is \(-\frac{\pi}{2}\), and plot the sine wave, noting that at every whole integer multiple of the period the function cycles. This is a sine wave with an amplitude of \(\frac{1}{2}\) (meaning it reaches a maximum height of \(\frac{1}{2}\) and minimum of \(-\frac{1}{2}\)), a period of 1, and shifted right by \(\frac{\pi}{2}\) units.