Problem 47 - Precalculus Essentials

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

Step 1: Identify the amplitude

The amplitude of a trigonometric function is the coefficient in front of it. In the given function, the amplitude is \(\frac{1}{2}\).

Step 2: Identify the period

The period of a cosine function is normally \(2 \pi\), but it can be varied depending on the coefficient of \(x\), given by \(\frac{2\pi}{|B|}\), where \(B\) is the coefficient of \(x\). In our function, the coefficient of \(x\) is 3. So, the period of this function is \(\frac{2 \pi}{3}\).

Step 3: Identify the phase shift

The phase shift of the function is given by \(-\frac{C}{B}\), where \(C\) is the constant inside the cosine, and \(B\) is the coefficient of \(x\). The given function has \(C = \frac{\pi}{2}\) and \(B = 3\), so the phase shift is \(-\frac{\frac{\pi}{2}}{3} = -\frac{\pi}{6}\), which means the function is shifted \(\frac{\pi}{6}\) units to the right.

Step 4: Draw the graph

Plot the graph of the function using the properties. The graph starts at the phase shift, reaches a maximum or a minimum (depending on the positive or negative direction of the function) at quarter period, returns to normal at half period, reaches its minimal or maximal point at the three quarter period and then starts a new period at the end of the full period. Repeat this pattern for one full period. The amplitude and period identified previously will guide the exact positioning of these points.