Answer :
To determine all the real numbers [tex]\( s \)[/tex] associated with the point [tex]\((x, y) = \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\)[/tex] on the unit circle, follow these steps:
### Step 1: Understand the Unit Circle
A point on the unit circle represents the cosine and sine values of an angle [tex]\(\theta\)[/tex]:
[tex]\[ (x, y) = (\cos(\theta), \sin(\theta)) \][/tex]
Given:
[tex]\[ x = -\frac{\sqrt{3}}{2} \quad \text{and} \quad y = \frac{1}{2} \][/tex]
### Step 2: Find the Principal Angle
To find the principal angle [tex]\(\theta\)[/tex] corresponding to the given point, we look at inverse trigonometric functions. Specifically, we can use the [tex]\(\text{arctan2}\)[/tex] function, which correctly handles the quadrant of the angle. The point [tex]\(\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\)[/tex] is in the second quadrant.
The principal angle [tex]\(\theta\)[/tex] in the range [tex]\([0, 2\pi)\)[/tex] is found to be:
[tex]\[ \theta \approx 2.6179938779914944 \text{ radians} \][/tex]
This means:
[tex]\[ \theta = \frac{5\pi}{6} \][/tex]
### Step 3: General Solution Using Periodicity
The unit circle is periodic with a period of [tex]\(2\pi\)[/tex]. Therefore, the general solution for the angle [tex]\(s\)[/tex] considering the periodic nature of trigonometric functions is given by:
[tex]\[ s = \theta + 2k\pi \quad \text{for any integer} \; k \][/tex]
Thus, the general solution is:
[tex]\[ s = \frac{5\pi}{6} + 2k\pi \][/tex]
where [tex]\( k \)[/tex] is any integer.
### Step 4: Consolidate the Answers
Hence, the exact answer within [tex]\([0, 2\pi)\)[/tex] is:
[tex]\[ s = \frac{5\pi}{6} \][/tex]
And the remaining solutions can be expressed using the periodicity:
[tex]\[ s = \frac{5\pi}{6} + 2k\pi \quad \text{where} \; k \; \text{is any integer} \][/tex]
By following these detailed steps, we have determined the principal angle and the general solution for all real numbers [tex]\(s\)[/tex] associated with the point [tex]\(\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\)[/tex] on the unit circle.
### Step 1: Understand the Unit Circle
A point on the unit circle represents the cosine and sine values of an angle [tex]\(\theta\)[/tex]:
[tex]\[ (x, y) = (\cos(\theta), \sin(\theta)) \][/tex]
Given:
[tex]\[ x = -\frac{\sqrt{3}}{2} \quad \text{and} \quad y = \frac{1}{2} \][/tex]
### Step 2: Find the Principal Angle
To find the principal angle [tex]\(\theta\)[/tex] corresponding to the given point, we look at inverse trigonometric functions. Specifically, we can use the [tex]\(\text{arctan2}\)[/tex] function, which correctly handles the quadrant of the angle. The point [tex]\(\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\)[/tex] is in the second quadrant.
The principal angle [tex]\(\theta\)[/tex] in the range [tex]\([0, 2\pi)\)[/tex] is found to be:
[tex]\[ \theta \approx 2.6179938779914944 \text{ radians} \][/tex]
This means:
[tex]\[ \theta = \frac{5\pi}{6} \][/tex]
### Step 3: General Solution Using Periodicity
The unit circle is periodic with a period of [tex]\(2\pi\)[/tex]. Therefore, the general solution for the angle [tex]\(s\)[/tex] considering the periodic nature of trigonometric functions is given by:
[tex]\[ s = \theta + 2k\pi \quad \text{for any integer} \; k \][/tex]
Thus, the general solution is:
[tex]\[ s = \frac{5\pi}{6} + 2k\pi \][/tex]
where [tex]\( k \)[/tex] is any integer.
### Step 4: Consolidate the Answers
Hence, the exact answer within [tex]\([0, 2\pi)\)[/tex] is:
[tex]\[ s = \frac{5\pi}{6} \][/tex]
And the remaining solutions can be expressed using the periodicity:
[tex]\[ s = \frac{5\pi}{6} + 2k\pi \quad \text{where} \; k \; \text{is any integer} \][/tex]
By following these detailed steps, we have determined the principal angle and the general solution for all real numbers [tex]\(s\)[/tex] associated with the point [tex]\(\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\)[/tex] on the unit circle.
