a. Determine a sine function that is defined for all *x*≥0 and has its first minimum at ( pi/3,3)

first maximum at (4𝜋/3,9)
b. State an equivalent cosine function for part a.

To determine a sine function that has its first minimum at \((\pi/3, 3)\) and its first maximum at \((4\pi/3, 9)\), let's start by analyzing the general form of a sine function:

\[ y = A \sin(Bx + C) + D \]

where:
- \(A\) is the amplitude,
- \(B\) affects the period of the function,
- \(C\) is the phase shift,
- \(D\) is the vertical shift.

Given the points:
- Minimum at \((\pi/3, 3)\)
- Maximum at \((4\pi/3, 9)\)

### Step-by-Step Solution:

#### 1. **Determine the Vertical Shift (\(D\))**:
The average of the maximum and minimum values gives the vertical shift:
\[ D = \frac{9 + 3}{2} = 6 \]

#### 2. **Determine the Amplitude (\(A\))**:
The amplitude is half the distance between the maximum and minimum values:
\[ A = \frac{9 - 3}{2} = 3 \]

#### 3. **Determine the Period and \(B\)**:
The sine function completes one full cycle between the minimum and the next minimum (or maximum to maximum). From \(\pi/3\) to \(4\pi/3\) (distance of \(4\pi/3 - \pi/3 = \pi\)) is half the period. Hence, the full period \(T\) is \(2\pi\):
\[ T = 2 \times (\frac{4\pi/3 - \pi/3}{1/2}) = 2\pi \]
Thus,
\[ B = \frac{2\pi}{T} = 1 \]

#### 4. **Determine the Phase Shift (\(C\))**:
Since the function has its first minimum at \((\pi/3, 3)\), we need to shift the basic sine function accordingly. A sine function has its minimum at \(3\pi/2\) (without any shifts). Thus, we solve for \(C\) in:
\[ B \left(\frac{\pi}{3}\right) + C = \frac{3\pi}{2} \]
\[ \frac{\pi}{3} + C = \frac{3\pi}{2} \]
\[ C = \frac{3\pi}{2} - \frac{\pi}{3} \]
\[ C = \frac{9\pi - 2\pi}{6} \]
\[ C = \frac{7\pi}{6} \]

Therefore, the sine function is:
\[ y = 3 \sin(x - \frac{7\pi}{6}) + 6 \]

### Equivalent Cosine Function:
A cosine function can be written in the form:
\[ y = A \cos(Bx + C) + D \]

The cosine function can be converted from the sine function using the phase shift:
\[ \cos(x) = \sin(x + \frac{\pi}{2}) \]

So the equivalent cosine function is:
\[ y = 3 \cos(x - \frac{7\pi}{6} - \frac{\pi}{2}) + 6 \]
\[ y = 3 \cos(x - \frac{7\pi + 3\pi}{6}) + 6 \]
\[ y = 3 \cos(x - \frac{10\pi}{6}) + 6 \]
\[ y = 3 \cos(x - \frac{5\pi}{3}) + 6 \]

Thus, the cosine function is:
\[ y = 3 \cos(x - \frac{5\pi}{3}) + 6 \]

### Summary:
a. The sine function is: 
\[ y = 3 \sin(x - \frac{7\pi}{6}) + 6 \]

b. The equivalent cosine function is:
\[ y = 3 \cos(x - \frac{5\pi}{3}) + 6 \]

Would you like any further details or have any questions?

Here are some related questions for you to consider:

1. How do you determine the phase shift in a trigonometric function?
2. What is the significance of the amplitude in a sine or cosine function?
3. How do you find the period of a sine or cosine function?
4. Can you explain the difference between a sine and a cosine function in terms of their graphs?
5. How does the vertical shift affect the graph of a sine or cosine function?
6. What happens to the graph of a sine function if you change the amplitude?
7. How do you convert a sine function to a cosine function and vice versa?
8. What is the impact of the phase shift on the starting point of a sine or cosine wave?

**Tip:** Remember that the phase shift in a sine or cosine function shifts the graph horizontally, changing where the function starts its cycle.

Learn how to determine a sine function that is defined for all x ≥ 0, with its first minimum at (π/3, 3) and first maximum at (4π/3, 9). Also, find the equivalent cosine function. This problem involves trigonometric functions, amplitude determination, phase shift calculation, and is suitable for high school students.

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