Question

which trigonometric function is represented in the graph below?
options:

• $\cos(\theta + \frac{\pi}{3}) + \frac{1}{2}$
• $\frac{1}{2}\sin(\frac{1}{2}(\theta + \frac{\pi}{6})) + 1$
• $\frac{1}{2}\sin(2(\theta - \frac{\pi}{12})) + \frac{1}{2}$
• $\frac{1}{2}\cos(2(\theta + \frac{\pi}{6})) + 1$

Explanation:

Step1: Analyze the general form of trigonometric functions

The general form of a sinusoidal function is \( y = A\sin(B(\theta - C)) + D \) or \( y = A\cos(B(\theta - C)) + D \), where:
- \( |A| \) is the amplitude,
- \( \frac{2\pi}{|B|} \) is the period,
- \( C \) is the phase shift,
- \( D \) is the vertical shift.

Step2: Determine the amplitude, period, vertical shift from the graph

- **Amplitude**: The distance from the midline to the peak. From the graph, the amplitude \( |A|=\frac{1}{2} \) (since the distance between the maximum and minimum is 1, so amplitude is \( \frac{1}{2} \)).
- **Period**: The length of one full cycle. Observing the graph, the period seems to be \( \pi \) (since a standard sine or cosine has period \( 2\pi \), and if the period here is \( \pi \), then \( \frac{2\pi}{|B|}=\pi \implies |B| = 2 \)).
- **Vertical shift**: The midline. The midline is at \( y=\frac{1}{2} \) or \( y = 1 \)? Wait, looking at the options, let's check the vertical shift. Let's re - evaluate. Wait, the third option has \( D=\frac{1}{2} \), the fourth has \( D = 1 \), the second has \( D = 1 \), the first has \( D=\frac{1}{2} \).

Wait, let's check the period again. If \( B = 2 \), period \( T=\frac{2\pi}{2}=\pi \). Let's check the phase shift and the function type.

Let's analyze each option:

Option 1: \( \cos(\theta+\frac{\pi}{3})+\frac{1}{2} \)

- Amplitude \( A = 1 \) (since the coefficient of cosine is 1), but our graph has amplitude \( \frac{1}{2} \), so eliminate this.

Option 2: \( \frac{1}{2}\sin(\frac{1}{2}(\theta+\frac{\pi}{6})) + 1 \)

- The coefficient of \( \theta \) inside the sine is \( \frac{1}{2} \), so period \( T=\frac{2\pi}{\frac{1}{2}} = 4\pi \), which is too long. Our graph has a shorter period, so eliminate.

Option 3: \( \frac{1}{2}\sin(2(\theta-\frac{\pi}{12}))+\frac{1}{2} \)

- Amplitude \( A=\frac{1}{2} \), period \( T=\frac{2\pi}{2}=\pi \), vertical shift \( D = \frac{2}{2}=\frac{1}{2} \) (wait, no, \( D=\frac{1}{2} \)). Let's check the phase shift. The phase shift is \( C=\frac{\pi}{12} \) to the right. Let's see the graph: the sine function shifted appropriately.

Option 4: \( \frac{1}{2}\cos(2(\theta+\frac{\pi}{6}))+1 \)

- Vertical shift \( D = 1 \), but our midline seems to be at \( y=\frac{1}{2} \) (since the amplitude is \( \frac{1}{2} \), if midline is \( \frac{1}{2} \), then max is \( \frac{1}{2}+\frac{1}{2}=1 \), min is \( \frac{1}{2}-\frac{1}{2}=0 \)). Wait, maybe I made a mistake in vertical shift. Wait, if the midline is \( \frac{1}{2} \), then vertical shift \( D=\frac{1}{2} \). Option 4 has \( D = 1 \), so midline at \( y = 1 \), which does not match.

Wait, let's re - check the third option: \( y=\frac{1}{2}\sin(2(\theta-\frac{\pi}{12}))+\frac{1}{2} \)

Expand it: \( y=\frac{1}{2}\sin(2\theta-\frac{\pi}{6})+\frac{1}{2} \)

The general form of sine is \( y = A\sin(B\theta - C)+D \), here \( A=\frac{1}{2}, B = 2, C=\frac{\pi}{6}, D=\frac{1}{2} \)

Amplitude \( \frac{1}{2} \), period \( \pi \), phase shift \( \frac{\pi}{12} \) to the right, vertical shift \( \frac{1}{2} \). This matches the amplitude, period, and vertical shift.

The fourth option: \( y=\frac{1}{2}\cos(2(\theta+\frac{\pi}{6})) + 1=\frac{1}{2}\cos(2\theta+\frac{\pi}{3})+1 \)

Amplitude \( \frac{1}{2} \), period \( \pi \), but vertical shift \( D = 1 \), which would make the midline at \( y = 1 \), but our graph's midline is at \( y=\frac{1}{2} \) (since amplitude is \( \frac{1}{2} \), so max - min = 1, midline is \( \frac{\max+\min}{2} \). If max is \( \frac{1}{2}+\frac{1}{2}=1 \), min is \( \frac{1}{2}-\frac{1}{2}=0 \), midline is \( \frac{1 + 0}{2}=\frac{1}{2} \)), so vertical shift \( D=\frac{1}{2} \), so option 4 is out.

So the correct option is the third one: \( \frac{1}{2}\sin(2(\theta-\frac{\pi}{12}))+\frac{1}{2} \)

Answer:

\( \boldsymbol{\frac{1}{2}\sin(2(\theta - \frac{\pi}{12}))+\frac{1}{2}} \) (the third option: \( \frac{1}{2}\sin(2(\theta-\frac{\pi}{12}))+\frac{1}{2} \))