Problem 47 - Algebra and Trigonometry

Question

Determine the amplitude, period, and phase shift of each function. Then graph one period of the function. $$y=\frac{1}{2} \cos \left(3 x+\frac{\pi}{2}\right)$$

Step-by-Step Solution

Answer

The amplitude is 1/2, the period is \( \frac{2\pi}{3} \) and the phase shift is - \( \frac{\pi}{6} \).

Step 1: Determine the Amplitude

The amplitude of the function is the absolute value of the coefficient of the cosine function. Here, the amplitude is the absolute value of 1/2, i.e., \( A = |1/2| = 1/2. \)

Step 2: Determine the Period

The period of a function is given by \( P = \frac{2\pi}{|B|} \), where \( B \) is the coefficient of \( x \) in the function. Here, the coefficient of \( x \) is 3. So, the period is \( P = \frac{2\pi}{|3|} = \frac{2\pi}{3}. \)

Step 3: Determine the Phase Shift

The phase shift of a function is given by \( S = -\frac{C}{B} \), where \( C \) is the horizontal translation of the function and \( B \) is the coefficient of \( x \). Here, \( C = \frac{\pi}{2} \) and \( B = 3 \). So, the phase shift is \( S = -\frac{\pi/2}{3} = -\frac{\pi}{6}. \)

Step 4: Graphing the Function

To graph the function, start by marking the amplitude on the vertical (y) axis and a full period on the horizontal (x) axis. The graph will oscillate between the amplitude, with one full wave completed in one period. Due to the phase shift, the graph will start from this value on the x-axis rather than at the origin.