Math Problem Statement

Find the reference number for each value of t. (a)
t = 4πœ‹ 7

(b)
t = βˆ’ 7πœ‹ 9

(c)
t = βˆ’3

(d)
t = 5

Solution

The reference number (also called the reference angle) is the positive acute angle that tt makes with the x-axis in the unit circle. It is always between 00 and Ο€2\frac{\pi}{2}.

Finding the reference number for each given value of tt:

(a) t=4Ο€7t = \frac{4\pi}{7}

  1. Determine the quadrant:
    • Since 4Ο€7\frac{4\pi}{7} is between 00 and Ο€\pi, it is in Quadrant II because Ο€=7Ο€7\pi = \frac{7\pi}{7}.
  2. Find the reference number: Reference number=Ο€βˆ’4Ο€7=7Ο€7βˆ’4Ο€7=3Ο€7\text{Reference number} = \pi - \frac{4\pi}{7} = \frac{7\pi}{7} - \frac{4\pi}{7} = \frac{3\pi}{7}

(b) t=βˆ’7Ο€9t = -\frac{7\pi}{9}

  1. Determine the quadrant:
    • Since tt is negative, we move clockwise.
    • βˆ’7Ο€9-\frac{7\pi}{9} is between βˆ’Ο€-\pi and 00, so it is in Quadrant III because βˆ’Ο€=βˆ’9Ο€9-\pi = -\frac{9\pi}{9}.
  2. Find the reference number: Reference number=βˆ£Ο€βˆ’(βˆ’7Ο€9)∣=∣9Ο€9βˆ’7Ο€9∣=2Ο€9\text{Reference number} = \left| \pi - \left(-\frac{7\pi}{9}\right) \right| = \left| \frac{9\pi}{9} - \frac{7\pi}{9} \right| = \frac{2\pi}{9}

(c) t=βˆ’3t = -3

  1. Convert to radians:
    • Ο€β‰ˆ3.14\pi \approx 3.14, so βˆ’3-3 is close to βˆ’Ο€-\pi.
    • Since βˆ’3-3 is in Quadrant III (because βˆ’Ο€<βˆ’3<0-\pi < -3 < 0), we compute the reference number:
  2. Find the reference number: Reference number=βˆ£βˆ’Ο€βˆ’(βˆ’3)∣=βˆ£Ο€βˆ’3∣\text{Reference number} = | -\pi - (-3) | = | \pi - 3 | Approximating Ο€β‰ˆ3.14\pi \approx 3.14, Reference numberβ‰ˆβˆ£3.14βˆ’3∣=0.14\text{Reference number} \approx |3.14 - 3| = 0.14

(d) t=5t = 5

  1. Convert to radians:
    • Since Ο€β‰ˆ3.14\pi \approx 3.14, we note that 5>Ο€5 > \pi, meaning it is in Quadrant IV.
    • The closest multiple of 2Ο€2\pi (which is about 6.28) is greater than 5, so it remains in the first rotation.
  2. Find the reference number: Reference number=∣5βˆ’Ο€βˆ£\text{Reference number} = |5 - \pi| Approximating Ο€β‰ˆ3.14\pi \approx 3.14, Reference numberβ‰ˆβˆ£5βˆ’3.14∣=1.86\text{Reference number} \approx |5 - 3.14| = 1.86

Final Answers:

(a) 3Ο€7\frac{3\pi}{7}
(b) 2Ο€9\frac{2\pi}{9}
(c) βˆ£Ο€βˆ’3βˆ£β‰ˆ0.14|\pi - 3| \approx 0.14
(d) ∣5βˆ’Ο€βˆ£β‰ˆ1.86|5 - \pi| \approx 1.86

Do you want a step-by-step breakdown for any part?

Related Questions:

  1. What is the general formula for finding reference numbers in different quadrants?
  2. How do reference numbers help in trigonometry?
  3. Can reference numbers be negative?
  4. How do reference numbers relate to periodic functions like sine and cosine?
  5. What is the significance of the unit circle in calculating reference numbers?

Tip: If the angle is negative, always move clockwise to find its quadrant.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Unit Circle
Reference Angle
Radians

Formulas

Reference Number = |Ο€ - t| (if in Quadrant II or III)
Reference Number = |t - Ο€| (if in Quadrant IV)
Reference Number = |Ο€ - 3| (for non-exact radian values)
Reference Number = |5 - Ο€| (for non-exact radian values)

Theorems

Reference Number Theorem
Quadrant Identification in Trigonometry
Unit Circle Properties

Suitable Grade Level

Grades 10-12