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To determine the terminal point on the unit circle for [tex]\(\frac{5\pi}{6}\)[/tex] radians, we first identify the position of the angle and then find its corresponding [tex]\((x, y)\)[/tex] coordinates.
1. Understanding the Angle:
- The angle [tex]\(\frac{5\pi}{6}\)[/tex] is in radians and converted to degrees, it is [tex]\(\frac{5}{6} \times 180^\circ = 150^\circ\)[/tex].
- This angle lies in the second quadrant since it is between [tex]\(90^\circ\)[/tex] and [tex]\(180^\circ\)[/tex].
2. Unit Circle Coordinates:
- For [tex]\(\theta = \frac{5\pi}{6}\)[/tex], the reference angle is found by subtracting from [tex]\(\pi\)[/tex]:
[tex]\[ \pi - \frac{5\pi}{6} = \frac{\pi}{6} \][/tex]
- The coordinates for the reference angle [tex]\(\frac{\pi}{6}\)[/tex] are [tex]\(\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\)[/tex], based on the properties of the unit circle where:
[tex]\[ \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \][/tex]
3. Adjusting for the Quadrant:
- In the second quadrant, the cosine (x-coordinate) is negative, and the sine (y-coordinate) is positive.
- Therefore, the coordinates for [tex]\(\theta = \frac{5\pi}{6}\)[/tex] would be:
[tex]\[ \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right) \][/tex]
4. Validation:
- Cosine of [tex]\(\frac{5\pi}{6}\)[/tex] is:
[tex]\[ \cos \left(\frac{5\pi}{6}\right) = -\cos \left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} \][/tex]
- Sine of [tex]\(\frac{5\pi}{6}\)[/tex] is:
[tex]\[ \sin \left(\frac{5\pi}{6}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \][/tex]
Thus, the terminal point on the unit circle for [tex]\(\frac{5\pi}{6}\)[/tex] radians is:
[tex]\[ \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right) \][/tex]
Therefore, the correct choice is:
[tex]\[ \boxed{\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)} \][/tex]
1. Understanding the Angle:
- The angle [tex]\(\frac{5\pi}{6}\)[/tex] is in radians and converted to degrees, it is [tex]\(\frac{5}{6} \times 180^\circ = 150^\circ\)[/tex].
- This angle lies in the second quadrant since it is between [tex]\(90^\circ\)[/tex] and [tex]\(180^\circ\)[/tex].
2. Unit Circle Coordinates:
- For [tex]\(\theta = \frac{5\pi}{6}\)[/tex], the reference angle is found by subtracting from [tex]\(\pi\)[/tex]:
[tex]\[ \pi - \frac{5\pi}{6} = \frac{\pi}{6} \][/tex]
- The coordinates for the reference angle [tex]\(\frac{\pi}{6}\)[/tex] are [tex]\(\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\)[/tex], based on the properties of the unit circle where:
[tex]\[ \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \][/tex]
3. Adjusting for the Quadrant:
- In the second quadrant, the cosine (x-coordinate) is negative, and the sine (y-coordinate) is positive.
- Therefore, the coordinates for [tex]\(\theta = \frac{5\pi}{6}\)[/tex] would be:
[tex]\[ \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right) \][/tex]
4. Validation:
- Cosine of [tex]\(\frac{5\pi}{6}\)[/tex] is:
[tex]\[ \cos \left(\frac{5\pi}{6}\right) = -\cos \left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} \][/tex]
- Sine of [tex]\(\frac{5\pi}{6}\)[/tex] is:
[tex]\[ \sin \left(\frac{5\pi}{6}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} \][/tex]
Thus, the terminal point on the unit circle for [tex]\(\frac{5\pi}{6}\)[/tex] radians is:
[tex]\[ \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right) \][/tex]
Therefore, the correct choice is:
[tex]\[ \boxed{\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)} \][/tex]
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