Exercise Set 4.3: Unit Circle Trigonometry Worksheet for 9th ... - Free Printable

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Problem 1: Match each of the following angles to standard position. Then use a protractor (and draw a quick sketch) of each angle.

This part asks you to sketch angles in standard position (vertex at origin, initial side on positive x-axis). The angles are:

1. $ \frac{\pi}{3} $ → 60°
2. $ \frac{5\pi}{4} $ → 225°
3. $ -\frac{\pi}{6} $ → -30° (clockwise)
4. $ \frac{7\pi}{6} $ → 210°
5. $ -\frac{3\pi}{4} $ → -135°
6. $ \frac{11\pi}{6} $ → 330°

> ✔ Explanation: To sketch these:
> - Convert radians to degrees if needed.
> - Draw the initial side along the positive x-axis.
> - Rotate counterclockwise for positive angles, clockwise for negative.
> - Use a protractor to estimate the angle.

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Problem 2: Match each of the following angles to standard position. Then use the unit circle to find the exact coordinates of the terminal point.

Let’s solve each:

#### a) $ \theta = \frac{\pi}{3} $
- Reference angle: $ \frac{\pi}{3} $
- In quadrant I
- Coordinates: $ \left( \cos\frac{\pi}{3}, \sin\frac{\pi}{3} \right) = \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) $

#### b) $ \theta = \frac{5\pi}{4} $
- $ \frac{5\pi}{4} = \pi + \frac{\pi}{4} $ → third quadrant
- Reference angle: $ \frac{\pi}{4} $
- Cosine and sine both negative
- $ \left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) $

#### c) $ \theta = -\frac{\pi}{6} $
- Equivalent to $ 360^\circ - 30^\circ = 330^\circ $
- Fourth quadrant
- $ \left( \cos(-\frac{\pi}{6}), \sin(-\frac{\pi}{6}) \right) = \left( \frac{\sqrt{3}}{2}, -\frac{1}{2} \right) $

#### d) $ \theta = \frac{7\pi}{6} $
- $ \pi + \frac{\pi}{6} $ → third quadrant
- Reference angle: $ \frac{\pi}{6} $
- Both sin and cos negative
- $ \left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right) $

#### e) $ \theta = -\frac{3\pi}{4} $
- Equivalent to $ 360^\circ - 135^\circ = 225^\circ $ → same as $ \frac{5\pi}{4} $
- Third quadrant
- $ \left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) $

#### f) $ \theta = \frac{11\pi}{6} $
- $ 2\pi - \frac{\pi}{6} = 330^\circ $
- Fourth quadrant
- $ \left( \frac{\sqrt{3}}{2}, -\frac{1}{2} \right) $

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Problem 3: Find three angles, one negative and two positive, that are coterminal with each of the following.

Coterminal angles differ by multiples of $ 2\pi $. So:

#### a) $ \frac{\pi}{4} $
- Add $ 2\pi $: $ \frac{\pi}{4} + 2\pi = \frac{9\pi}{4} $
- Add $ 4\pi $: $ \frac{\pi}{4} + 4\pi = \frac{17\pi}{4} $
- Subtract $ 2\pi $: $ \frac{\pi}{4} - 2\pi = -\frac{7\pi}{4} $

→ Answers: $ \frac{9\pi}{4}, \frac{17\pi}{4}, -\frac{7\pi}{4} $

#### b) $ \frac{3\pi}{2} $
- $ +2\pi $: $ \frac{3\pi}{2} + 2\pi = \frac{7\pi}{2} $
- $ +4\pi $: $ \frac{11\pi}{2} $
- $ -2\pi $: $ \frac{3\pi}{2} - 2\pi = -\frac{\pi}{2} $

→ Answers: $ \frac{7\pi}{2}, \frac{11\pi}{2}, -\frac{\pi}{2} $

#### c) $ \frac{5\pi}{6} $
- $ +2\pi $: $ \frac{5\pi}{6} + 2\pi = \frac{17\pi}{6} $
- $ +4\pi $: $ \frac{29\pi}{6} $
- $ -2\pi $: $ \frac{5\pi}{6} - 2\pi = -\frac{7\pi}{6} $

→ Answers: $ \frac{17\pi}{6}, \frac{29\pi}{6}, -\frac{7\pi}{6} $

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Problem 4: Solve the following

#### a) Use the trigonometric ratios to determine the values of $ \sin \theta $, $ \cos \theta $, and $ \tan \theta $ for $ \theta = \frac{\pi}{4} $

- $ \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} $
- $ \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2} $
- $ \tan\frac{\pi}{4} = 1 $

#### b) Use the trigonometric ratios to determine the values of $ \sin \theta $, $ \cos \theta $, and $ \tan \theta $ for $ \theta = \frac{5\pi}{3} $

- $ \frac{5\pi}{3} = 300^\circ $ → fourth quadrant
- Reference angle: $ \frac{\pi}{3} $
- $ \sin\frac{5\pi}{3} = -\frac{\sqrt{3}}{2} $
- $ \cos\frac{5\pi}{3} = \frac{1}{2} $
- $ \tan\frac{5\pi}{3} = \frac{-\sqrt{3}/2}{1/2} = -\sqrt{3} $

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Problem 5: Use the reference angle to evaluate the following.

#### a) $ \sin\left(\frac{4\pi}{3}\right) $
- $ \frac{4\pi}{3} = \pi + \frac{\pi}{3} $ → third quadrant
- Reference angle: $ \frac{\pi}{3} $
- Sine is negative in QIII
- $ \sin\left(\frac{4\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} $

#### b) $ \cos\left(\frac{7\pi}{6}\right) $
- $ \frac{7\pi}{6} = \pi + \frac{\pi}{6} $ → third quadrant
- Reference angle: $ \frac{\pi}{6} $
- Cosine is negative in QIII
- $ \cos\left(\frac{7\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} $

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Problem 6: Using the following unit circle, draw and then label the terminal side of an angle $ \theta $ such that:

#### a) $ \sin \theta = \frac{1}{2} $
- Sine is y-coordinate
- On unit circle, $ y = \frac{1}{2} $ at:
- $ \theta = \frac{\pi}{6} $ (QI)
- $ \theta = \frac{5\pi}{6} $ (QII)
- Draw terminal side at either of those positions.

#### b) $ \cos \theta = -\frac{\sqrt{3}}{2} $
- Cosine is x-coordinate
- $ x = -\frac{\sqrt{3}}{2} $ occurs at:
- $ \theta = \frac{5\pi}{6} $ (QII)
- $ \theta = \frac{7\pi}{6} $ (QIII)
- Draw terminal side at either.

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Problem 7: Use the given information to determine the value of $ \theta $ in the interval $ [0, 2\pi) $.

#### a) $ \sin \theta = \frac{1}{2} $, $ \cos \theta > 0 $
- Sine is positive in QI and QII
- Cosine is positive in QI and QIV
- So only QI satisfies both
- $ \theta = \frac{\pi}{6} $

#### b) $ \tan \theta = -1 $, $ \sin \theta < 0 $
- Tangent is negative in QII and QIV
- Sine is negative in QIII and QIV
- So only QIV satisfies both
- In QIV, $ \tan \theta = -1 $ → $ \theta = \frac{7\pi}{4} $

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✔ Final Summary of Key Answers:

| Problem | Answer |
|--------|--------|
| 2a | $ \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) $ |
| 2b | $ \left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) $ |
| 2c | $ \left( \frac{\sqrt{3}}{2}, -\frac{1}{2} \right) $ |
| 2d | $ \left( -\frac{\sqrt{3}}{2}, -\frac{1}{2} \right) $ |
| 2e | $ \left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) $ |
| 2f | $ \left( \frac{\sqrt{3}}{2}, -\frac{1}{2} \right) $ |
| 3a | $ \frac{9\pi}{4}, \frac{17\pi}{4}, -\frac{7\pi}{4} $ |
| 3b | $ \frac{7\pi}{2}, \frac{11\pi}{2}, -\frac{\pi}{2} $ |
| 3c | $ \frac{17\pi}{6}, \frac{29\pi}{6}, -\frac{7\pi}{6} $ |
| 4a | $ \sin = \frac{\sqrt{2}}{2}, \cos = \frac{\sqrt{2}}{2}, \tan = 1 $ |
| 4b | $ \sin = -\frac{\sqrt{3}}{2}, \cos = \frac{1}{2}, \tan = -\sqrt{3} $ |
| 5a | $ -\frac{\sqrt{3}}{2} $ |
| 5b | $ -\frac{\sqrt{3}}{2} $ |
| 7a | $ \frac{\pi}{6} $ |
| 7b | $ \frac{7\pi}{4} $ |

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If you’d like me to explain any particular concept (like reference angles, unit circle, coterminal angles), let me know!