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Let’s go through the problem and analyze it.

The unit circle diagram you provided shows an angle \( \theta = \frac{5\pi}{6} \), and you are being asked if the statement by Josiah, who claims that \( \sin \left( \frac{5\pi}{6} \right) = -\frac{\sqrt{3}}{2} \), is correct.

### Steps to Verify:
1. **Location on the Unit Circle:**
   - The angle \( \frac{5\pi}{6} \) lies in the second quadrant.
   - In this quadrant, sine is positive, and cosine is negative.

2. **Sine of \( \frac{5\pi}{6} \):**
   - \( \sin \left( \frac{5\pi}{6} \right) \) has the same value as \( \sin \left( \frac{\pi}{6} \right) \), because \( \frac{5\pi}{6} \) and \( \frac{\pi}{6} \) are supplementary angles (they add up to \( \pi \)).
   - \( \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \).
   - So, \( \sin \left( \frac{5\pi}{6} \right) = \frac{1}{2} \), and **not** \( -\frac{\sqrt{3}}{2} \).

3. **Conclusion:**
   Josiah's claim is incorrect because the sine of \( \frac{5\pi}{6} \) is \( \frac{1}{2} \), not \( -\frac{\sqrt{3}}{2} \).

If this clears things up, feel free to let me know! Would you like additional details or explanations?

### Related Questions:
1. Why is the sine function positive in the second quadrant?
2. How can you find the cosine of \( \frac{5\pi}{6} \)?
3. What is the general method to find sine or cosine for angles on the unit circle?
4. Can sine ever be negative for angles between \( \pi \) and \( 2\pi \)?
5. How does the unit circle help in determining trigonometric functions for different angles?

### Tip:
For angles in the unit circle, remember that the sign of sine and cosine depends on the quadrant. In the second quadrant, sine is positive and cosine is negative!

Based on the unit circle shown, Josiah claims that sin(5π/6) = -√3/2. Is Josiah correct?

Explore whether Josiah's claim that sin(5π/6) = -√3/2 is correct by using the unit circle and understanding the sine function behavior in different quadrants. Suitable for students in Grades 10-12.

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