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Find the terminal point on the unit circle determined by [tex]\frac{11 \pi}{6}[/tex] radians. Use exact values, not decimal approximations.

A. [tex]\left(-\frac{1}{2}, \frac{\sqrt{2}}{2}\right)[/tex]
B. [tex]\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)[/tex]
C. [tex]\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)[/tex]
D. [tex]\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)[/tex]


Sagot :

To determine the terminal point on the unit circle for the angle [tex]\(\frac{11 \pi}{6}\)[/tex] radians, we need to compute the corresponding coordinates [tex]\((x, y)\)[/tex] where [tex]\(x = \cos(\theta)\)[/tex] and [tex]\(y = \sin(\theta)\)[/tex].

### Step-by-Step Solution:

1. Identify the angle:
[tex]\[ \theta = \frac{11 \pi}{6} \][/tex]

2. Locate the quadrant:
- The angle [tex]\(\frac{11 \pi}{6}\)[/tex] is just slightly less than [tex]\(2\pi\)[/tex] (since [tex]\(2\pi = \frac{12 \pi}{6}\)[/tex]).
- Since [tex]\(2\pi\)[/tex] represents a full circle (360 degrees), [tex]\(\frac{11 \pi}{6}\)[/tex] is located in the fourth quadrant.

3. Reference angle:
- The reference angle for angles in the fourth quadrant is calculated as:
[tex]\[ 2\pi - \frac{11 \pi}{6} = \frac{12 \pi}{6} - \frac{11 \pi}{6} = \frac{\pi}{6} \][/tex]
- So, the reference angle is [tex]\(\frac{\pi}{6}\)[/tex] radians.

4. Exact values for sine and cosine:
- From trigonometric identities, we know the exact values for the sine and cosine of [tex]\(\frac{\pi}{6}\)[/tex]:
[tex]\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \][/tex]
[tex]\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \][/tex]

5. Sign adjustments in the fourth quadrant:
- In the fourth quadrant:
- [tex]\(x\)[/tex] (cosine) is positive.
- [tex]\(y\)[/tex] (sine) is negative.

Thus:
[tex]\[ \cos\left(\frac{11 \pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \][/tex]
[tex]\[ \sin\left(\frac{11 \pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2} \][/tex]

6. Final coordinates:
[tex]\[ \left( \cos\left(\frac{11 \pi}{6}\right), \sin\left(\frac{11 \pi}{6}\right) \right) = \left( \frac{\sqrt{3}}{2}, -\frac{1}{2} \right) \][/tex]

Given the choices:
- [tex]\(\left(-\frac{1}{2}, \frac{\sqrt{2}}{2}\right)\)[/tex]
- [tex]\(\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)\)[/tex]
- [tex]\(\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\)[/tex]
- [tex]\(\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)\)[/tex]

The correct terminal point on the unit circle determined by [tex]\(\frac{11 \pi}{6}\)[/tex] radians is:
[tex]\[ \boxed{\left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)} \][/tex]
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