For the following values of t find (a) the reference number of t and (b) the terminal point determined
by t.
i. t = 7π/
3
ii. t = − 11π/
6
iii. t = 17π/4

To solve these problems, we'll find both the reference number and the terminal point on the unit circle for each value of \( t \). 

### Understanding Reference Number and Terminal Point
1. **Reference Number**: It is the acute angle between the terminal side of the given angle \( t \) and the x-axis. It helps in simplifying trigonometric calculations.
2. **Terminal Point**: It is the point on the unit circle corresponding to the angle \( t \).

### Part (i): \( t = \frac{7\pi}{3} \)

1. **Convert \( t \) to an equivalent angle between \( 0 \) and \( 2\pi \):**
   \[
   \frac{7\pi}{3} = 2\pi + \frac{\pi}{3}
   \]
   So, the equivalent angle within \( 0 \) to \( 2\pi \) is \( \frac{\pi}{3} \).

2. **Reference Number**: 
   Since \( \frac{\pi}{3} \) is already in the first quadrant, the reference number is \( \frac{\pi}{3} \).

3. **Terminal Point**:
   The terminal point for \( \frac{\pi}{3} \) is:
   \[
   \left( \cos\left(\frac{\pi}{3}\right), \sin\left(\frac{\pi}{3}\right) \right) = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right).
   \]

### Part (ii): \( t = -\frac{11\pi}{6} \)

1. **Convert \( t \) to an equivalent positive angle:**
   \[
   -\frac{11\pi}{6} + 2\pi = -\frac{11\pi}{6} + \frac{12\pi}{6} = \frac{\pi}{6}.
   \]

2. **Reference Number**: 
   The reference number is \( \frac{\pi}{6} \), as the angle is in the first quadrant.

3. **Terminal Point**:
   The terminal point for \( \frac{\pi}{6} \) is:
   \[
   \left( \cos\left(\frac{\pi}{6}\right), \sin\left(\frac{\pi}{6}\right) \right) = \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right).
   \]

### Part (iii): \( t = \frac{17\pi}{4} \)

1. **Convert \( t \) to an equivalent angle between \( 0 \) and \( 2\pi \):**
   \[
   \frac{17\pi}{4} = 4\pi + \frac{\pi}{4} = 2\pi + 2\pi + \frac{\pi}{4}.
   \]
   So, the equivalent angle is \( \frac{\pi}{4} \).

2. **Reference Number**: 
   Since \( \frac{\pi}{4} \) is in the first quadrant, the reference number is \( \frac{\pi}{4} \).

3. **Terminal Point**:
   The terminal point for \( \frac{\pi}{4} \) is:
   \[
   \left( \cos\left(\frac{\pi}{4}\right), \sin\left(\frac{\pi}{4}\right) \right) = \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right).
   \]

### Summary

1. \( t = \frac{7\pi}{3} \):
   - Reference Number: \( \frac{\pi}{3} \)
   - Terminal Point: \( \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \)

2. \( t = -\frac{11\pi}{6} \):
   - Reference Number: \( \frac{\pi}{6} \)
   - Terminal Point: \( \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) \)

3. \( t = \frac{17\pi}{4} \):
   - Reference Number: \( \frac{\pi}{4} \)
   - Terminal Point: \( \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) \)

Would you like more details or have any questions?

### Related Questions:
1. How do you find the reference angle for negative values of \( t \)?
2. What is the relationship between reference angles and their corresponding trigonometric values?
3. How does adding \( 2\pi \) affect the terminal point?
4. Can a reference number ever be greater than \( \pi/2 \)?
5. How do terminal points help in understanding the unit circle?

### Tip:
Always convert angles to a standard range (usually \( 0 \) to \( 2\pi \)) to easily identify the reference angle and terminal point.

For the following values of t find (a) the reference number of t and (b) the terminal point determined by t.
 i. t = 7π/3
 ii. t = −11π/6
 iii. t = 17π/4

Discover how to find the reference angle and terminal point for angles t = 7π/3, −11π/6, and 17π/4. This problem involves trigonometric concepts like the unit circle, reference angles, and terminal points, suitable for high school students in Grades 9-12.

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