1. **State the problem:** We are given a point on the unit circle with coordinates $$\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$$ and asked to find: i) The six trigonometric function values. ii) The quadrant where this point lies. iii) The angle made by this point with the positive x-axis. 2. **Calculate all six trigonometric functions:** - Sine is the y-coordinate: $$\sin \theta = \frac{1}{2}$$ - Cosine is the x-coordinate: $$\cos \theta = -\frac{\sqrt{3}}{2}$$ - Tangent is sine divided by cosine: $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$$ - Cotangent is the reciprocal of tangent: $$\cot \theta = \frac{1}{\tan \theta} = -\sqrt{3}$$ - Secant is the reciprocal of cosine: $$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$$ - Cosecant is the reciprocal of sine: $$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{1}{2}} = 2$$ 3. **Determine the quadrant:** - The x-coordinate is negative and the y-coordinate is positive. - Therefore, the point lies in the **second quadrant**, where x < 0 and y > 0. 4. **Calculate the angle formed:** - Recognize the reference angle where sine is \( \frac{1}{2} \) and cosine is \( \frac{\sqrt{3}}{2} \), which is \(30^\circ\) or \(\frac{\pi}{6}\). - Since the point is in the second quadrant, the angle with the positive x-axis is: $$\theta = 180^\circ - 30^\circ = 150^\circ$$ or in radians: $$\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$ **Final summary:** - $$\sin \theta = \frac{1}{2}, \quad \cos \theta = -\frac{\sqrt{3}}{2}, \quad \tan \theta = -\frac{\sqrt{3}}{3}$$ - $$\cot \theta = -\sqrt{3}, \quad \sec \theta = -\frac{2\sqrt{3}}{3}, \quad \csc \theta = 2$$ - Quadrant: Second - Angle: $$\theta = \frac{5\pi}{6}$$ radians or $$150^\circ$$