Question

Use the table to answer the question.
𝜃 𝑓(𝜃)
0 2
0
−2
0
2

Determine the missing inputs in your table of values for the first cycle of the function f(θ)=2cos(5θ/2).

Answer 1

To determine the missing values in the table for the function \( f(\theta) = 2\cos\left(\frac{5\theta}{2}\right) \), we first need to identify the values of \( \theta \) corresponding to the output values of \( f(\theta) \).

### Step 1: Identify Values of \( \theta \)
The function \( f(\theta) = 2\cos\left(\frac{5\theta}{2}\right) \) oscillates between -2 and 2. The interested \( \theta \) values to fill in the gaps are points where \( f(\theta) \) produces the known outputs (2, 0, -2) and completes a full cycle within the range of \( \theta \).

### Step 2: Calculate Outputs for Specific Inputs of \( \theta \)
To find the outputs of the function for key angles in the first cycle (from \( 0 \) to \( 2\pi \)), we find points where the \( \cos \) function takes on specific values:

1. **When \( f(\theta) = 2 \)**:
- \( 2\cos\left(\frac{5\theta}{2}\right) = 2 \)
- \( \cos\left(\frac{5\theta}{2}\right) = 1 \)
- This occurs when \( \frac{5\theta}{2} = 2k\pi \) for \( k \in \mathbb{Z} \).
- For \( k = 0 \): \( \frac{5\theta}{2} = 0 \Rightarrow \theta = 0 \) (corresponds to \( f(0) = 2 \))

2. **When \( f(\theta) = 0 \)**:
- \( 2\cos\left(\frac{5\theta}{2}\right) = 0 \)
- \( \cos\left(\frac{5\theta}{2}\right) = 0 \)
- This occurs when \( \frac{5\theta}{2} = \frac{\pi}{2} + k\pi \).
- For \( k = 0 \): \( \frac{5\theta}{2} = \frac{\pi}{2} \Rightarrow \theta = \frac{\pi}{5} \)
- For \( k = 1 \): \( \frac{5\theta}{2} = \frac{3\pi}{2} \Rightarrow \theta = \frac{3\pi}{5} \)

3. **When \( f(\theta) = -2 \)**:
- \( 2\cos\left(\frac{5\theta}{2}\right) = -2 \)
- \( \cos\left(\frac{5\theta}{2}\right) = -1 \)
- This occurs when \( \frac{5\theta}{2} = (2k + 1)\pi \).
- For \( k = 0 \): \( \frac{5\theta}{2} = \pi \Rightarrow \theta = \frac{2\pi}{5} \)

### Step 3: Summary and Filling in the Table
Using these outputs, we identify:
- \( \theta = 0 \) produces \( f(0) = 2 \)
- \( \theta = \frac{\pi}{5} \) produces \( f\left(\frac{\pi}{5}\right) = 0 \)
- \( \theta = \frac{2\pi}{5} \) produces \( f\left(\frac{2\pi}{5}\right) = -2 \)
- \( \theta = \frac{3\pi}{5} \) produces \( f\left(\frac{3\pi}{5}\right) = 0 \)
- \( \theta = \frac{4\pi}{5} \) produces \( f\left(\frac{4\pi}{5}\right) = 2 \)

Thus, the completed table looks like:

\[
\begin{array}{|c|c|}
\hline
\theta & f(\theta) \\
\hline
0 & 2 \\
\frac{\pi}{5} & 0 \\
\frac{2\pi}{5} & -2 \\
\frac{3\pi}{5} & 0 \\
\frac{4\pi}{5} & 2 \\
\hline
\end{array}
\]

These values capture the first cycle of the function \( f(\theta) \).